{"segment":[{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. {$1;\\dfrac{m+4}{m-1}$}","B. {$2;\\dfrac{m+4}{m+1}$}","C. {$1;\\dfrac{m+4}{m+1}$}"],"ques":"<span class='basic_left'>Nh\u1ea9m nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\left( m-1 \\right){{x}^{2}}-\\left( 2m+3 \\right)x+m+4=0$ v\u1edbi $m \\ne 1$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${?}","hint":"","explain":"<span class='basic_left'>V\u1edbi $m \\ne 1$ th\u00ec $m-1 \\ne 0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai.<br\/>Ta c\u00f3: $a+b+c=(m-1)-(2m+3)+m+4=0$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=1;{{x}_{2}}=\\dfrac{m+4}{m-1}$<\/span><\/span>"}]}],"id_ques":891},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $-{{x}^{2}}+m\\left( m+1 \\right)x-{{m}^{2}}+m+1=0$ c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t","select":["A. $m\\ne 1;m\\ne -2$","B. $m\\ne -1;m\\ne 2$","C. $m\\ge -1;m\\ne 2$","D. $-1 \\le m \\le 2$"],"hint":"T\u00ednh $a+b+c$ sau \u0111\u00f3 nh\u1ea9m nghi\u1ec7m","explain":"<span class='basic_left'>Do $a+b+c=-1+m(m+1)-m^2+m+1=0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}}=1;{{x}_{2}}={{m}^{2}}-m-1$ <br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t th\u00ec ${{x}_{1}}\\ne {{x}_{2}}$ <br\/>$\\begin{aligned} & \\Leftrightarrow 1\\ne {{m}^{2}}-m-1 \\\\ & \\Leftrightarrow {{m}^{2}}-m-2\\ne 0 \\\\ & \\Leftrightarrow \\left( m+1 \\right)\\left( m-2 \\right)\\ne 0 \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne -1 \\\\ & m\\ne 2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>V\u1eady v\u1edbi $m\\ne -1;m\\ne 2$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>C\u00e1ch 2: T\u00ednh $\\Delta$ r\u1ed3i t\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $m$ \u0111\u1ec3 $\\Delta >0$. Tuy nhi\u00ean \u1edf b\u00e0i n\u00e0y, t\u00ednh $\\Delta$ kh\u00e1 ph\u1ee9c t\u1ea1p, ta ph\u1ea3i gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc 4. C\u00e1c b\u1ea1n c\u00f3 th\u1ec3 ph\u00e2n t\u00edch th\u00e0nh b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh t\u00edch r\u1ed3i gi\u1ea3i.<\/span>","column":2}]}],"id_ques":892},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Bi\u1ebft ${{u}^{2}}+{{v}^{2}}=85,uv=18.$<br\/>C\u00f3 bao nhi\u00eau c\u1eb7p s\u1ed1 $(u;v)$ th\u1ecfa m\u00e3n b\u00e0i to\u00e1n? ","select":["A. 1","B. 2","C. 3","D. 4"],"hint":"T\u1eeb gi\u1ea3 thi\u1ebft, ta t\u00ecm t\u1ed5ng $u+v.$ ","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u1eeb hai h\u1ec7 th\u1ee9c \u0111\u00e3 cho, ta t\u00ednh t\u1ed5ng $S=u+v$<br\/>B\u01b0\u1edbc 2: Ki\u1ec3m tra c\u00f3 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n $S^2-4P \\ge 0$. N\u1ebfu th\u1ecfa m\u00e3n th\u00ec $u$ v\u00e0 $v$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: $x^2-Sx+P=0$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/> T\u1eeb ${{u}^{2}}+{{v}^{2}}=85,uv=18$, ta c\u00f3 ${{u}^{2}}+2uv+{{v}^{2}}=85+2.18=121$ <br\/>Suy ra ${{\\left( u+v \\right)}^{2}}={{11}^{2}}\\Leftrightarrow u+v=\\pm 11$ <br\/>+ Tr\u01b0\u1eddng h\u1ee3p 1: $\\left\\{ \\begin{align} & u+v=11 \\\\ & u.v=18 \\\\ \\end{align} \\right.$ <br\/>Ta th\u1ea5y $S^2-4P=11^2-4.18=49>0$ n\u00ean $u$ v\u00e0 $v$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: $x^2-11x+18=0$<br\/>$\\Delta =49$ n\u00ean ${{x}_{1}}=\\dfrac{11+7}{2}=9;{{x}_{2}}=\\dfrac{11-7}{2}=2$ <br\/>Suy ra $u=9;v=2$ ho\u1eb7c $u=2; v=9$<br\/>+ Tr\u01b0\u1eddng h\u1ee3p 2: $\\left\\{ \\begin{align} & u+v=-11 \\\\ & u.v=18 \\\\ \\end{align} \\right.$<br\/>Ta th\u1ea5y $S^2-4P=(-11)^2-4.18=49>0$ n\u00ean $u$ v\u00e0 $v$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: $x^2+11x+18=0$<br\/>$\\Delta =49$ n\u00ean ${{x}_{1}}=\\dfrac{-11+7}{2}=-2;{{x}_{2}}=\\dfrac{-11-7}{2}=-9$ <br\/>Suy ra $u=-9;v=-2$ ho\u1eb7c $u=-2; v=-9$<br\/>V\u1eady c\u00e1c c\u1eb7p s\u1ed1 $(u;v)$ th\u1ecfa m\u00e3n b\u00e0i to\u00e1n l\u00e0 $(9;2);(2;9);(-9;-2);(-2;-9)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span><\/span>","column":4}]}],"id_ques":893},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+mx-5=0$ c\u00f3 nghi\u1ec7m l\u00e0 ${{x}_{1}}$ v\u00e0 ${{x}_{2}}$ . Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 hai s\u1ed1 $\\dfrac{1}{{{x}_{1}}}$ v\u00e0 $\\dfrac{1}{{{x}_{2}}}$ l\u00e0: ","select":["A. $5{{x}^{2}}-x-m=0$","B. $5{{x}^{2}}-mx+1=0$","C. $5{{x}^{2}}+mx-1=0$","D. $5{{x}^{2}}-mx-1=0$"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, t\u00ednh t\u1ed5ng v\u00e0 t\u00edch hai nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+mx-5=0$<br\/>B\u01b0\u1edbc 2: \u0110\u1eb7t $u=\\dfrac{1}{{{x}_{1}}};v=\\dfrac{1}{{{x}_{2}}}$. T\u00ednh t\u1ed5ng $S=u+v$ v\u00e0 t\u00edch $P=uv$ <br\/> Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 hai nghi\u1ec7m $u;v$ l\u00e0 $x^2-Sx+P=0$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+mx-5=0$ c\u00f3 $\\Delta ={{m}^{2}}+20>0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}$ v\u00e0 ${{x}_{2}}$ ph\u00e2n bi\u1ec7t<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi \u2013\u00e9t, ta c\u00f3: $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=-m \\\\ & {{x}_{1}}.{{x}_{2}}=-5 \\\\ \\end{align} \\right.$<br\/> \u0110\u1eb7t $u=\\dfrac{1}{{{x}_{1}}};v=\\dfrac{1}{{{x}_{2}}}$ <br\/>Ta c\u00f3 $S=u+v=\\dfrac{1}{{{x}_{1}}}+\\dfrac{1}{{{x}_{2}}}=\\dfrac{{{x}_{1}}+{{x}_{2}}}{{{x}_{1}}.{{x}_{2}}}=\\dfrac{m}{5}$ ; $P=uv=\\dfrac{1}{{{x}_{1}}{{x}_{2}}}=-\\dfrac{1}{5}$<br\/> ${{S}^{2}}-4P={{\\left( \\dfrac{m}{5} \\right)}^{2}}+\\dfrac{4}{5}=\\dfrac{{{m}^{2}}+20}{25}>0$ <br\/>Khi \u0111\u00f3 $u$ v\u00e0 $v$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{2}}-\\dfrac{m}{5}x-\\dfrac{1}{5}=0$ hay $5{{x}^{2}}-mx-1=0$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 hai s\u1ed1 $\\dfrac{1}{{{x}_{1}}}$ v\u00e0 $\\dfrac{1}{{{x}_{2}}}$ l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh $5{{x}^{2}}-mx-1=0$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D <\/span> <\/span>","column":2}]}],"id_ques":894},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["5"],["-23"]]],"list":[{"point":5,"width":50,"content":"","type_input":"","ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+5x+3=0$ c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ . L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 hai nghi\u1ec7m l\u00e0 ${{y}_{1}}=2{{x}_{1}}-{{x}_{2}};{{y}_{2}}=2{{x}_{2}}-{{x}_{1}}$<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> Ph\u01b0\u01a1ng tr\u00ecnh l\u1eadp \u0111\u01b0\u1ee3c l\u00e0: $X^2+$_input_$X+$_input_$=0$<\/span> ","hint":"T\u00ednh t\u1ed5ng v\u00e0 t\u00edch hai s\u1ed1 $y_1$ v\u00e0 $y_2$ r\u1ed3i l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh","explain":"<span class='basic_left'>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi \u2013\u00e9t, ta c\u00f3: ${{x}_{1}}+{{x}_{2}}=-5;{{x}_{1}}.{{x}_{2}}=3$ <br\/>Ta c\u00f3<br\/> $\\begin{align} & S={{y}_{1}}+{{y}_{2}}=\\left( 2{{x}_{1}}-{{x}_{2}} \\right)+\\left( 2{{x}_{2}}-{{x}_{1}} \\right)={{x}_{1}}+{{x}_{2}}=-5; \\\\ & P={{y}_{1}}.{{y}_{2}}=\\left( 2{{x}_{1}}-{{x}_{2}} \\right).\\left( 2{{x}_{2}}-{{x}_{1}} \\right)=4{{x}_{1}}{{x}_{2}}-2x_{1}^{2}-2x_{2}^{2}+{{x}_{1}}{{x}_{2}} \\\\ & \\,\\,\\,\\,=5{{x}_{1}}{{x}_{2}}-2\\left[ {{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}} \\right]=15-2\\left( 25-6 \\right)=-23 \\\\ \\end{align}$<br\/> Ta c\u00f3: ${{S}^{2}}-4P=117>0.$ <br\/> Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 hai nghi\u1ec7m l\u00e0 $y_1$ v\u00e0 $y_2$ l\u00e0 $x^2+5x-23=0$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $5$ v\u00e0 $-23.$<\/span><\/span>"}]}],"id_ques":895},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Gi\u1ea3 s\u1eed ${{x}_{1}};{{x}_{2}}$ l\u00e0 hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+mx+1=0$ v\u1edbi $m$ l\u00e0 tham s\u1ed1.<br\/> T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a $A=\\dfrac{x_{1}^{2}}{x_{2}^{2}}+\\dfrac{x_{2}^{2}}{x_{1}^{2}}$ theo $m$","select":["A. $A={{m}^{4}}+4{{m}^{2}}+2$ ","B. $A={{m}^{4}}-4{{m}^{2}}+2$ "],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/> B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 nghi\u1ec7m: $\\Delta \\ge 0$<br\/>B\u01b0\u1edbc 2: T\u1eeb h\u1ec7 th\u1ee9c Vi \u2013 \u00e9t, t\u00ednh $S$ v\u00e0 $P$ (t\u1ed5ng v\u00e0 t\u00edch c\u00e1c nghi\u1ec7m) <br\/>B\u01b0\u1edbc 3: Bi\u1ec3u di\u1ec5n bi\u1ec3u th\u1ee9c \u0111\u1ed1i x\u1ee9ng qua $S$ v\u00e0 $P,$ sau \u0111\u00f3 thay $S$ v\u00e0 $P$ t\u00ednh \u0111\u01b0\u1ee3c \u1edf tr\u00ean v\u00e0o bi\u1ec3u th\u1ee9c \u0111\u1ed1i x\u1ee9ng.<br\/>Bi\u1ec3u th\u1ee9c gi\u1eefa ${{x}_{1}};{{x}_{2}}$ nh\u01b0 tr\u00ean g\u1ecdi l\u00e0 bi\u1ec3u th\u1ee9c \u0111\u1ed1i x\u1ee9ng: Khi ta thay ${{x}_{1}}$ b\u1edfi ${{x}_{2}}$ v\u00e0 ${{x}_{2}}$ b\u1edfi ${{x}_{1}}$ th\u00ec bi\u1ec3u th\u1ee9c kh\u00f4ng \u0111\u1ed5i.<br\/> <span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m $\\Leftrightarrow {{m}^{2}}-4\\ge 0\\Leftrightarrow \\left| m \\right|\\ge 2$ <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi \u2013 \u00e9t, ta c\u00f3 $\\left\\{ \\begin{align} & S={{x}_{1}}+{{x}_{2}}=-m \\\\ & P={{x}_{1}}{{x}_{2}}=1 \\\\ \\end{align} \\right.$ <br\/>Ta c\u00f3:<br\/> $\\begin{align} & A={{\\left( \\dfrac{{{x}_{1}}}{{{x}_{2}}}+\\dfrac{{{x}_{2}}}{{{x}_{1}}} \\right)}^{2}}-2.\\dfrac{{{x}_{1}}}{{{x}_{2}}}.\\dfrac{{{x}_{2}}}{{{x}_{1}}} \\\\ & \\,\\,\\,\\,\\,\\,={{\\left( \\dfrac{x_{1}^{2}+x_{2}^{2}}{{{x}_{1}}{{x}_{2}}} \\right)}^{2}}-2 \\\\ & \\,\\,\\,\\,\\,\\,={{\\left[ \\dfrac{{{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}}}{{{x}_{1}}{{x}_{2}}} \\right]}^{2}}-2 \\\\ & \\,\\,\\,\\,\\,\\,={{\\left( \\dfrac{{{S}^{2}}-2P}{P} \\right)}^{2}}-2 \\\\ & \\,\\,\\,\\,\\,\\,={{\\left( {{m}^{2}}-2 \\right)}^{2}}-2 \\\\ & \\,\\,\\,\\,\\,\\,={{m}^{4}}-4{{m}^{2}}+2 \\\\ \\end{align}$<br\/>V\u1eady $A={{m}^{4}}-4{{m}^{2}}+2$ <br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span> <br\/><span class='basic_green'>Ch\u00fa \u00fd:<\/span><br\/>\u0110\u1ec3 t\u00ednh \u0111\u01b0\u1ee3c t\u1ed5ng v\u00e0 t\u00edch hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh theo h\u1ec7 th\u1ee9c Vi \u2013 \u00e9t, ta c\u1ea7n ki\u1ec3m tra ho\u1eb7c t\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u00e0 c\u00f3 nghi\u1ec7m<\/span>","column":2}]}],"id_ques":896},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+2\\left( m+1 \\right)x+{{m}^{2}}+4m+3=0$ v\u1edbi $m$ l\u00e0 tham s\u1ed1. Gi\u1ea3 s\u1eed ${{x}_{1}};{{x}_{2}}$ l\u00e0 hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. \u0110\u1eb7t $A={{x}_{1}}{{x}_{2}}-2\\left( {{x}_{1}}+{{x}_{2}} \\right)$ .<br\/> Khi \u0111\u00f3 $A={{m}^{2}}+8m+7,$ <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?<\/span>","select":["A. \u0110\u00fang ","B. Sai"],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 $\\Delta '={{\\left( m+1 \\right)}^{2}}-\\left( {{m}^{2}}+4m+3 \\right)=-2m-2$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $\\Leftrightarrow \\Delta '\\ge 0\\Leftrightarrow m<-1$ <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi \u2013 \u00e9t, ta c\u00f3: $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=-2\\left( m+1 \\right) \\\\ & {{x}_{1}}.{{x}_{2}}={{m}^{2}}+4m+3 \\\\ \\end{align} \\right.$ <br\/>Suy ra<br\/> $\\begin{align} & A={{m}^{2}}+4m+3+4\\left( m+1 \\right) \\\\ & \\,\\,\\,\\,\\,={{m}^{2}}+4m+3+4m+4 \\\\ & \\,\\,\\,\\,\\,={{m}^{2}}+8m+7 \\\\ \\end{align}$ <br\/>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0111\u00fang.<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":897},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","f","f"]],"list":[{"point":5,"image":"img\/1.png","col_name":["","\u0110\u00fang","Sai"],"arr_ques":[" Hai ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0\\,\\,\\left( a\\ne 0 \\right)$ v\u00e0 $a{{x}^{2}}-bx+c=0\\,\\,\\left( a\\ne 0 \\right)$ c\u00f9ng c\u00f3 nghi\u1ec7m ho\u1eb7c c\u00f9ng v\u00f4 nghi\u1ec7m","Ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ v\u1edbi $a\\ne 0$ c\u00f3 hai nghi\u1ec7m c\u00f9ng d\u1ea5u khi t\u00edch hai nghi\u1ec7m d\u01b0\u01a1ng.","Ph\u01b0\u01a1ng tr\u00ecnh $\\left( m-1 \\right){{x}^{2}}+3mx+2m+1=0$ c\u00f3 hai nghi\u1ec7m l\u00e0 $-1$ v\u00e0 $\\dfrac{2m+1}{1-m}$ "],"hint":"","explain":["<span class='basic_left'>\u0110\u00fang v\u00ec c\u1ea3 hai ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec1u c\u00f3 $\\Delta ={{b}^{2}}-4ac$ n\u00ean c\u1ea3 hai ph\u01b0\u01a1ng tr\u00ecnh c\u00f9ng c\u00f3 nghi\u1ec7m ho\u1eb7c c\u00f9ng v\u00f4 nghi\u1ec7m<\/span>","<span class='basic_left'><br\/> Sai v\u00ec ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ v\u1edbi $a\\ne 0$ c\u00f3 hai nghi\u1ec7m c\u00f9ng d\u1ea5u khi v\u00e0 ch\u1ec9 khi $\\left\\{ \\begin{align} & \\Delta \\ge 0 \\\\ & P={{x}_{1}}{{x}_{2}}>0 \\\\ \\end{align} \\right.$<br\/> Do \u0111\u00f3 \u0111i\u1ec1u ki\u1ec7n t\u00edch hai nghi\u1ec7m d\u01b0\u01a1ng ch\u01b0a \u0111\u1ee7 \u0111\u1ec3 k\u1ebft lu\u1eadn ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m c\u00f9ng d\u1ea5u.<\/span>","<span class='basic_left'><br\/> Sai v\u00ec v\u1edbi $m=1$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t ch\u1ec9 c\u00f3 m\u1ed9t nghi\u1ec7m $x=-1$<\/span>"]}]}],"id_ques":898},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-6mx+8=0$ . T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ th\u1ecfa m\u00e3m $x_{1}^{3}+x_{2}^{3}=72$ <br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $m=$_input_","hint":"\u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(a^3+b^3)=(a+b)(a^2-ab+b^2)$ \u0111\u1ec3 bi\u1ebfn \u0111\u1ed5i h\u1ec7 th\u1ee9c gi\u1eefa $x_1$ v\u00e0 $x_2$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m<br\/>B\u01b0\u1edbc 2: \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi -\u00e9t \u0111\u1ec3 t\u00ednh t\u1ed5ng v\u00e0 t\u00edch hai nghi\u1ec7m<br\/>B\u01b0\u1edbc 3: Bi\u1ebfn \u0111\u1ed5i h\u1ec7 th\u1ee9c \u0111\u00e3 cho v\u1ec1 d\u1ea1ng ch\u1ec9 ch\u1ee9a t\u1ed5ng v\u00e0 t\u00edch hai nghi\u1ec7m. T\u1eeb \u0111\u00f3 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00ecm $m.$<br\/>B\u01b0\u1edbc 4: Ki\u1ec3m tra $m$ c\u00f3 th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n kh\u00f4ng r\u1ed3i k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-6mx+8=0$ c\u00f3 nghi\u1ec7m $\\Leftrightarrow \\Delta '=9{{m}^{2}}-8\\ge 0\\Leftrightarrow {{m}^{2}}\\ge \\dfrac{8}{9}$ (*)<br\/>Theo h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=6m \\\\ & {{x}_{1}}.{{x}_{2}}=8 \\\\ \\end{align} \\right.$ <br\/>Ta c\u00f3:<br\/>$\\begin{align} & x_{1}^{3}+x_{2}^{3}=\\left( {{x}_{1}}+{{x}_{2}} \\right)\\left( x_{1}^{2}-{{x}_{1}}{{x}_{2}}+x_{2}^{2} \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=\\left( {{x}_{1}}+{{x}_{2}} \\right)\\left[ {{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-3{{x}_{1}}{{x}_{2}} \\right] \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=6m\\left( 36{{m}^{2}}-24 \\right) \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,=72m\\left( 3{{m}^{2}}-2 \\right) \\\\ \\end{align}$ <br\/>Theo gi\u1ea3 thi\u1ebft, ta c\u00f3 <br\/>$\\begin{align} & x_{1}^{3}+x_{2}^{3}=72 \\\\ & \\Leftrightarrow 72m\\left( 3{{m}^{2}}-2 \\right)=72 \\\\ & \\Leftrightarrow 3{{m}^{3}}-2m-1=0 \\\\ & \\Leftrightarrow \\left( m-1 \\right)\\left( 3{{m}^{2}}+3m+1 \\right)=0\\,\\,\\,\\left( 1 \\right) \\\\ \\end{align}$<br\/>V\u00ec $3{{m}^{2}}+3m+1=3{{\\left( m+\\dfrac{1}{2} \\right)}^{2}}+\\dfrac{1}{4}>0$ v\u1edbi m\u1ecdi $m$ n\u00ean <br\/>$\\left( 1 \\right)\\Leftrightarrow m-1=0\\Leftrightarrow m=1$ (th\u1ecfa m\u00e3n (*))<br\/>V\u1eady $m=1$ th\u00ec $x_{1}^{3}+x_{2}^{3}=72$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1.$<\/span>"}]}],"id_ques":899},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["2"],["-2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3 s\u1eed ${{x}_{1}};{{x}_{2}}$ l\u00e0 hai nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+2mx+4=0$ . X\u00e1c \u0111\u1ecbnh $m$ \u0111\u1ec3 $x_{1}^{4}+x_{2}^{4}\\le 32$ <br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $m \\in $ {_input_; _input_}","hint":"","explain":"<span class='basic_left'>Ta c\u00f3 $\\Delta '={{m}^{2}}-4$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m n\u00ean $\\Delta '\\ge 0\\Leftrightarrow {{m}^{2}}\\ge 4\\Leftrightarrow \\left| m \\right|\\ge 2$ (1)<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $S={{x}_{1}}+{{x}_{2}}=-2m;P={{x}_{1}}.{{x}_{2}}=4$<br\/>$\\begin{align} & x_{1}^{4}+x_{2}^{4}\\le 32 \\\\ & \\Leftrightarrow {{\\left( x_{1}^{2}+x_{2}^{2} \\right)}^{2}}-2x_{1}^{2}x_{2}^{2}\\le 32 \\\\ & \\Leftrightarrow {{\\left[ {{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}} \\right]}^{2}}-2x_{1}^{2}x_{2}^{2}\\le 32 \\\\ & \\Leftrightarrow {{\\left( {{S}^{2}}-2P \\right)}^{2}}-2{{P}^{2}}\\le 32 \\\\ & \\Leftrightarrow {{\\left( 4{{m}^{2}}-8 \\right)}^{2}}-32\\le 32 \\\\ & \\Leftrightarrow {{\\left( 4{{m}^{2}}-8 \\right)}^{2}}\\le 64 \\\\ & \\Leftrightarrow 16{{m}^{4}}-64{{m}^{2}}\\le 0 \\\\ & \\Leftrightarrow 16{{m}^{2}}\\left( {{m}^{2}}-4 \\right)\\le 0 \\\\ & \\Leftrightarrow {{m}^{2}}-4\\le 0\\,\\,\\,\\,\\,\\left( \\text{v\u00ec}\\,{{m}^{2}}\\ge 0 \\right) \\\\ & \\Leftrightarrow \\left| m \\right|\\le 2 \\\\ \\end{align}$<br\/>K\u1ebft h\u1ee3p v\u1edbi (1), ta c\u00f3 $m=\\pm 2$ <br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$ v\u00e0 $-2$<\/span>"}]}],"id_ques":900},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho ph\u01b0\u01a1ng tr\u00ecnh: $\\left( m-4 \\right){{x}^{2}}-2\\left( m-2 \\right)x+m-1=0$ . T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u v\u00e0 nghi\u1ec7m \u00e2m c\u00f3 gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i l\u1edbn h\u01a1n nghi\u1ec7m d\u01b0\u01a1ng.","select":["A. $ 1 < m < 4 $","B. $ 2 < m < 4 $","C. $ m > 4 $ ho\u1eb7c $ m < 2 $","D. $ m > 4 $ ho\u1eb7c $ m < 1 $"],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u $ \\Leftrightarrow ac<0 \\Leftrightarrow \\left( m-4 \\right) \\left( m-1 \\right)<0 \\Leftrightarrow 1 < m < 4 $ (1)<br\/>Gi\u1ea3 s\u1eed ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ m\u00e0 ${{x}_{1}}>0>{{x}_{2}}$ .<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $x_1+x_2=\\dfrac{2\\left( m-2 \\right)}{m-4}$<br\/>Nghi\u1ec7m \u00e2m c\u00f3 gi\u00e1 tr\u1ecb tuy\u1ec7t \u0111\u1ed1i l\u1edbn h\u01a1n nghi\u1ec7m d\u01b0\u01a1ng n\u00ean ta c\u00f3 <br\/>$\\left| {{x}_{2}} \\right|>{{x}_{1}}\\Leftrightarrow -{{x}_{2}}>{{x}_{1}}\\Leftrightarrow {{x}_{1}}+{{x}_{2}}<0\\Leftrightarrow \\dfrac{2\\left( m-2 \\right)}{m-4}<0\\Leftrightarrow 2 < m < 4 $ (2)<br\/>T\u1eeb (1) v\u00e0 (2), ta c\u00f3 $ 2 < m < 4$ th\u00ec th\u1ecfa m\u00e3n \u0111\u1ec1 b\u00e0i <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":901},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Gi\u1ea3 s\u1eed ${{x}_{1}};{{x}_{2}}$ l\u00e0 nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-2\\left( m-1 \\right)x+{{m}^{2}}-1=0$ . H\u1ec7 th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa ${{x}_{1}}$ v\u00e0 ${{x}_{2}}$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o $m$ l\u00e0:","select":["A. $2{{x}_{1}}{{x}_{2}}={{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2\\left( {{x}_{1}}+{{x}_{2}} \\right)$","B. $4{{x}_{1}}{{x}_{2}}={{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}+2\\left( {{x}_{1}}+{{x}_{2}} \\right)$","C. $4{{x}_{1}}{{x}_{2}}={{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}+4\\left( {{x}_{1}}+{{x}_{2}} \\right)$","D. $4{{x}_{1}}{{x}_{2}}={{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-4\\left( {{x}_{1}}+{{x}_{2}} \\right)$"],"hint":"T\u1eeb h\u1ec7 th\u1ee9c Vi-\u00e9t, ta bi\u1ec3u di\u1ec5n $m$ theo t\u1ed5ng $S$ v\u00e0 t\u00edch $P$ r\u1ed3i suy ra h\u1ec7 th\u1ee9c gi\u1eefa $x_1$ v\u00e0 $x_2$ kh\u00f4ng ph\u1ee5 thu\u1ed9c $m$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ecm \u0111i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 nghi\u1ec7m: $\\Delta \\ge 0$ <br\/>B\u01b0\u1edbc 2: T\u1eeb h\u1ec7 th\u1ee9c Vi \u2013 \u00e9t, t\u00ecm $S$ v\u00e0 $P$ theo tham s\u1ed1 $m.$<br\/>B\u01b0\u1edbc 3: Kh\u1eed tham s\u1ed1 m t\u1eeb $S$ v\u00e0 $P$ \u0111\u1ec3 c\u00f3 h\u1ec7 th\u1ee9c gi\u1eefa $S$ v\u00e0 $P$ kh\u00f4ng ph\u1ee5 thu\u1ed9c $m.$ \u0110\u00f3 ch\u00ecnh l\u00e0 h\u1ec7 th\u1ee9c gi\u1eefa ${{x}_{1}};{{x}_{2}}$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o $m.$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3 $\\Delta '={{\\left( m-1 \\right)}^{2}}-\\left( {{m}^{2}}-1 \\right)=m_2-2m+1-m^2+1=-2m+2$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m $\\Leftrightarrow \\Delta '\\ge 0\\Leftrightarrow -2m+2\\ge 0\\Leftrightarrow m\\le 1$ <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi -\u00e9t, ta c\u00f3: $\\left\\{ \\begin{align} & S={{x}_{1}}+{{x}_{2}}=2\\left( m-1 \\right)\\,\\,\\,\\,\\left( 1 \\right) \\\\ & P={{x}_{1}}{{x}_{2}}={{m}^{2}}-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align} \\right.$ <br\/>T\u1eeb (1) suy ra $m=\\dfrac{S+2}{2}$ . Thay v\u00e0o (2), ta \u0111\u01b0\u1ee3c:<br\/>$\\begin{align} & P={{\\left( \\dfrac{S+2}{2} \\right)}^{2}}-1 \\\\ & \\Leftrightarrow 4P={{S}^{2}}+4S \\\\ & \\Leftrightarrow 4{{x}_{1}}{{x}_{2}}={{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}+4\\left( {{x}_{1}}+{{x}_{2}} \\right) \\\\ \\end{align}$ <br\/>V\u1eady h\u1ec7 th\u1ee9c c\u1ea7n t\u00ecm l\u00e0 $4{{x}_{1}}{{x}_{2}}={{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}+4\\left( {{x}_{1}}+{{x}_{2}} \\right)$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":902},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Gi\u1ea3 s\u1eed ${{x}_{1}};{{x}_{2}}$ l\u00e0 hai nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh $m{{x}^{2}}-2\\left( m+1 \\right)x+m-4=0.$ H\u1ec7 th\u1ee9c li\u00ean h\u1ec7 gi\u1eefa ${{x}_{1}}$ v\u00e0 ${{x}_{2}}$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o $m$ l\u00e0:","select":["A. ${{x}_{1}}{{x}_{2}}=5+2\\left( {{x}_{1}}+{{x}_{2}} \\right)$","B. ${{x}_{1}}{{x}_{2}}=5-2\\left( {{x}_{1}}+{{x}_{2}} \\right)$","C. ${{x}_{1}}{{x}_{2}}=5-3\\left( {{x}_{1}}+{{x}_{2}} \\right)$","D. ${{x}_{1}}{{x}_{2}}=3+2\\left( {{x}_{1}}+{{x}_{2}} \\right)$"],"hint":"T\u1eeb h\u1ec7 th\u1ee9c Vi-\u00e9t, ta bi\u1ec3u di\u1ec5n $m$ theo t\u1ed5ng v\u00e0 t\u00edch hai nghi\u1ec7m r\u1ed3i suy ra h\u1ec7 th\u1ee9c gi\u1eefa $x_1$ v\u00e0 $x_2$ kh\u00f4ng ph\u1ee5 thu\u1ed9c $m$","explain":"<span class='basic_left'>Ta c\u00f3 $\\Delta '={{\\left( m+1 \\right)}^{2}}-m\\left( m-4 \\right)=m^2+2m+1-m^2+4m=6m+1$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m $ \\Leftrightarrow \\left \\{ \\begin{aligned} & a \\ne 0 \\\\ & \\Delta '\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & 6m+1\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & m \\ge -\\dfrac{1}{6} \\\\ \\end{aligned} \\right.$ <br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: <br\/>$\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=\\dfrac{2\\left( m+1 \\right)}{m}\\,\\,\\, \\\\ & {{x}_{1}}{{x}_{2}}=\\dfrac{m-4}{m}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=2+\\dfrac{2}{m} \\\\ & {{x}_{1}}{{x}_{2}}=1-\\dfrac{4}{m} \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & \\dfrac{2}{m}={{x}_{1}}+{{x}_{2}}-2 \\\\ & {{x}_{1}}{{x}_{2}}=1-\\dfrac{4}{m} \\\\ \\end{aligned} \\right.$ <br\/>Thay $\\dfrac{2}{m}={{x}_{1}}+{{x}_{2}}-2$ v\u00e0o h\u1ec7 th\u1ee9c th\u1ee9 hai, ta c\u00f3: <br\/> $\\begin{align} & {{x}_{1}}{{x}_{2}}=1-2\\left( {{x}_{1}}+{{x}_{2}}-2 \\right) \\\\ & \\Leftrightarrow {{x}_{1}}{{x}_{2}}=5-2\\left( {{x}_{1}}+{{x}_{2}} \\right) \\\\ \\end{align}$ <br\/>V\u1eady h\u1ec7 th\u1ee9c c\u1ea7n t\u00ecm l\u00e0 ${{x}_{1}}{{x}_{2}}=5-2\\left( {{x}_{1}}+{{x}_{2}} \\right)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B <\/span><\/span>","column":2}]}],"id_ques":903},{"time":24,"part":[{"title":"\u0110i\u1ec1n c\u1eb7p s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"V\u00ed d\u1ee5 b\u1ea1n t\u00ecm \u0111\u01b0\u1ee3c nghi\u1ec7m l\u00e0 $(2;3)$ th\u00ec b\u1ea1n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $(2;3).$<br\/> Ch\u00fa \u00fd: \u0110i\u1ec1n c\u1ea3 d\u1ea5u ngo\u1eb7c v\u00e0 d\u1ea5u ch\u1ea5m ph\u1ea9y.","temp":"fill_the_blank_random","correct":[[["(1;2)"],["(2;1)"]]],"list":[{"point":5,"width":80,"content":"","type_input":"","ques":"Gi\u1ea3i h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{aligned} & x+y+xy=5 \\\\ & {{x}^{2}}+{{y}^{2}}=5 \\\\ \\end{aligned} \\right.$ <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m $(x;y)$ l\u00e0 _input_ v\u00e0 _input_","hint":"\u0110\u1eb7t $S=x+y;P=xy.$ \u0110\u01b0a h\u1ec7 \u0111\u00e3 cho v\u1ec1 h\u1ec7 m\u1edbi v\u1edbi hai \u1ea9n $S$ v\u00e0 $P$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: \u0110\u1eb7t $S=x+y;P=xy.$ \u0110\u01b0a h\u1ec7 \u0111\u00e3 cho v\u1ec1 h\u1ec7 m\u1edbi v\u1edbi hai \u1ea9n $S$ v\u00e0 $P$<br\/>B\u01b0\u1edbc 2: Gi\u1ea3i h\u1ec7 t\u00ecm $S$ v\u00e0 $P.$ Khi \u0111\u00f3 $x, y$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ${{X}^{2}}-SX+P=0$ (\u0111i\u1ec1u ki\u1ec7n: ${{S}^{2}}\\ge 4P$ )<br\/>B\u01b0\u1edbc 3: N\u1ebfu $(x;y)$ l\u00e0 nghi\u1ec7m th\u00ec $(y;x)$ c\u0169ng l\u00e0 nghi\u1ec7m. K\u1ebft lu\u1eadn nghi\u1ec7m c\u1ee7a h\u1ec7.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ta c\u00f3:<br\/>$\\left\\{ \\begin{aligned} & x+y+xy=5 \\\\ & {{x}^{2}}+{{y}^{2}}=5 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x+y+xy=5 \\\\ & {{\\left( x+y \\right)}^{2}}-2xy=5 \\\\ \\end{aligned} \\right.$<br\/>\u0110\u1eb7t $S=x+y;P=xy$. Ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\left\\{ \\begin{aligned} & S+P=5\\,\\, \\\\ & {{S}^{2}}-2P=5 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & P=5-S\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & {{S}^{2}}-2P=5\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{aligned} \\right.$ <br\/>Thay (1) v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (2), ta c\u00f3: <br\/>$\\begin{aligned} & {{S}^{2}}-2\\left( 5-S \\right)=5 \\\\ & \\Leftrightarrow {{S}^{2}}+2S-15=0 \\\\ \\end{aligned}$ <br\/>$\\Delta '=1+15=16$ <br\/>Suy ra $\\left[ \\begin{aligned} & S=3 \\\\ & S=-5 \\\\ \\end{aligned} \\right.$ <br\/>+ V\u1edbi $S=3,$ ta c\u00f3 $P=5-3=2$ (th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n ${{S}^{2}}\\ge 4P$ ). Khi \u0111\u00f3 $x,y$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh:<br\/>${{X}^{2}}-3X+2=0\\Leftrightarrow \\left[ \\begin{aligned} & X=1 \\\\ & X=2 \\\\ \\end{aligned} \\right.$ (do $a+b+c=0$) <br\/>Suy ra $(x;y) =(1;2)$ ho\u1eb7c $(x;y)=(2;1)$<br\/>+ V\u1edbi $S=-5;P=5-(-5)=10$ (lo\u1ea1i v\u00ec kh\u00f4ng th\u1ecfa m\u00e3n ${{S}^{2}}\\ge 4P$).<br\/>V\u1eady h\u1ec7 \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m l\u00e0 $(1;2)$v\u00e0 $(2;1)$<br\/><span class='basic_pink'>V\u1eady c\u1eb7p s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $(1;2)$ v\u00e0 $(2;1).$<\/span>"}]}],"id_ques":904},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Cho ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-\\left( m+5 \\right)x+3m+6=0$ v\u1edbi $m$ l\u00e0 tham s\u1ed1. T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ${{x}_{1}};{{x}_{2}}$ l\u00e0 \u0111\u1ed9 d\u00e0i hai c\u1ea1nh g\u00f3c vu\u00f4ng c\u1ee7a m\u1ed9t tam gi\u00e1c vu\u00f4ng c\u00f3 c\u1ea1nh huy\u1ec1n b\u1eb1ng $5.$<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m=$_input_","hint":"\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Vi-\u00e9t v\u00e0 \u0111\u1ecbnh l\u00ed Pytago trong tam gi\u00e1c vu\u00f4ng","explain":"<span class='basic_left'>Ta c\u00f3 <br\/>$\\begin{aligned} & \\Delta ={{\\left( m+5 \\right)}^{2}}-4\\left( 3m+6 \\right) \\\\ & \\,\\,\\,\\,\\,={{m}^{2}}+10m+25-12m-24 \\\\ & \\,\\,\\,\\,\\,={{m}^{2}}-2m+1 \\\\ & \\,\\,\\,\\,={{\\left( m-1 \\right)}^{2}}\\ge 0\\,\\,\\,\\,\\forall m \\\\ \\end{aligned}$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m v\u1edbi m\u1ecdi $m$<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00ed Vi \u2013 \u00e9t, ta c\u00f3: $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=m+5 \\\\ & {{x}_{1}}{{x}_{2}}=3m+6 \\\\ \\end{aligned} \\right.$<br\/>V\u00ec hai nghi\u1ec7m $x_1$ v\u00e0 $x_2$ c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \u0111\u1ed9 d\u00e0i hai c\u1ea1nh g\u00f3c vu\u00f4ng n\u00ean $x_1>0$ v\u00e0 $x_2>0$. Khi \u0111\u00f3<br\/>$\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{x}}>0 \\\\ & {{x}_{1}}.{{x}_{2}}>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m+5>0 \\\\ & 3m+6>0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m>-5 \\\\ & m>-2 \\\\ \\end{aligned} \\right.\\Leftrightarrow m>-2$ <br\/>L\u1ea1i c\u00f3: ${{x}_{1}};{{x}_{2}}$ l\u00e0 \u0111\u1ed9 d\u00e0i hai c\u1ea1nh g\u00f3c vu\u00f4ng c\u1ee7a m\u1ed9t tam gi\u00e1c vu\u00f4ng c\u00f3 c\u1ea1nh huy\u1ec1n b\u1eb1ng $5.$<br\/> \u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3:<br\/>$\\begin{aligned} & x_{1}^{2}+x_{2}^{2}=25 \\\\ & \\Leftrightarrow {{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}}=25 \\\\ & \\Leftrightarrow {{\\left( m+5 \\right)}^{2}}-2\\left( 3m+6 \\right)=25 \\\\ & \\Leftrightarrow {{m}^{2}}+10m+25-6m-12=25 \\\\ & \\Leftrightarrow {{m}^{2}}+4m-12=0 \\\\ & \\Leftrightarrow \\left( m-2 \\right)\\left( m+6 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & m=2\\,\\text{(th\u1ecfa m\u00e3n)} \\\\ & m=-6 \\,(\\text{lo\u1ea1i v\u00ec kh\u00f4ng th\u1ecfa m\u00e3n}\\,m>-2) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>V\u1eady $m=2$ th\u00ec th\u1ecfa m\u00e3n y\u00eau c\u1ea7u b\u00e0i to\u00e1n.<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2.$<\/span>"}]}],"id_ques":905},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":" T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $\\left( m-1 \\right){{x}^{2}}-2\\left( m+1 \\right)x+m+2=0$ c\u00f3 hai nghi\u1ec7m \u0111\u1ed1i nhau.","select":["A. $m=-2$","B. $ m\\ge -3 $ v\u00e0 $ m \\ne 1 $ ","C. $ -2 < m < 1 $","D. $m=-1$"],"hint":"Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m \u0111\u1ed1i nhau th\u00ec hai nghi\u1ec7m \u0111\u00f3 tr\u00e1i d\u1ea5u v\u00e0 c\u00f3 t\u1ed5ng b\u1eb1ng $0.$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & \\Delta '={{\\left( m+1 \\right)}^{2}}-\\left( m-1 \\right)\\left( m+2 \\right) \\\\ & \\,\\,\\,\\,\\,\\,={{m}^{2}}+2m+1-\\left( {{m}^{2}}+m-2 \\right) \\\\ & \\,\\,\\,\\,\\,\\,=m+3 \\\\ \\end{aligned}$ <br\/>\u0110i\u1ec1u ki\u1ec7n \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0<br\/> $\\left\\{ \\begin{aligned} & m-1\\ne 0 \\\\ & \\Delta '\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 1 \\\\ & m+3\\ge 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 1 \\\\ & m\\ge -3 \\\\ \\end{aligned} \\right.\\,\\,\\,\\left( * \\right)$<br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi \u2013 \u00e9t, ta c\u00f3: $\\left\\{ \\begin{aligned} & {{x}_{1}}+{{x}_{2}}=\\dfrac{2\\left( m+1 \\right)}{m-1} \\\\ & {{x}_{1}}.{{x}_{2}}=\\dfrac{m+2}{m-1} \\\\ \\end{aligned} \\right.$<br\/> Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m \u0111\u1ed1i nhau $\\Leftrightarrow \\left\\{ \\begin{align} & {{x}_{1}}{{x}_{2}}<0 \\\\ & {{x}_{1}}+{{x}_{2}}=0 \\\\ \\end{align} \\right.$ <br\/>$\\Leftrightarrow \\left\\{ \\begin{align} & \\dfrac{m+2}{m-1}<0 \\\\ & \\dfrac{2\\left( m+1 \\right)}{m-1}=0 \\\\ \\end{align} \\right.$ $ \\Leftrightarrow \\left\\{ \\begin{aligned} & -2 < m < 1 \\\\ & m=-1 \\\\ \\end{aligned} \\right.$ $ \\Leftrightarrow m=-1 $ (th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n (*))<br\/>V\u1eady $m=-1$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m \u0111\u1ed1i nhau<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":906},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $\\left( m+1 \\right){{x}^{2}}-3mx+4m=0$ c\u00f3 nghi\u1ec7m ${{x}_{2}}>{{x}_{1}}>0$ <\/span>","select":["A. $ -\\dfrac{16}{7} < m < -1 $","B. $ -\\dfrac{16}{7} < m < 0 $ ","C. $ m > 1 $ ho\u1eb7c $ m < -\\dfrac{16}{7}$","D. $ 0 < m < \\dfrac{16}{7}$ "],"hint":"Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m ${{x}_{2}}>{{x}_{1}}>0,$ t\u1ee9c ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t","explain":"<span class='basic_left'>Ta c\u00f3<br\/>$\\Delta =9{{m}^{2}}-16m\\left( m+1 \\right)=9{{m}^{2}}-16{{m}^{2}}-16m=-m\\left( 7m+16 \\right)$ <br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m ${{x}_{2}}>{{x}_{1}}>0$ (t\u1ee9c hai nghi\u1ec7m d\u01b0\u01a1ng ph\u00e2n bi\u1ec7t) c\u1ea7n c\u00e1c \u0111i\u1ec1u ki\u1ec7n:$\\left\\{ \\begin{aligned} & a\\ne 0 \\\\ & \\Delta >0 \\\\ & S>0 \\\\ & P>0 \\\\ \\end{aligned} \\right.$ <br\/>+ $a\\ne 0\\Leftrightarrow m+1\\ne 0\\Leftrightarrow m\\ne -1$ <br\/>+ $ \\Delta > 0$ $\\Leftrightarrow -m\\left( 7m+16 \\right) > 0$ $\\Leftrightarrow \\left[ \\begin{aligned} & -m > 0;7m+16 > 0 \\\\ & -m < 0;7m+16 < 0 \\\\ \\end{aligned} \\right.$$\\Leftrightarrow \\left[ \\begin{aligned} & -\\dfrac{16}{7}< m < 0 \\\\ & 0 < m <-\\dfrac{16}{7}\\,\\text{(v\u00f4 l\u00ed)} \\\\ \\end{aligned} \\right.$$\\Leftrightarrow -\\dfrac{16}{7} < m < 0 $ <br\/>+ $ S > 0 \\Leftrightarrow \\dfrac{3m}{m+1} > 0$$\\Leftrightarrow \\left[ \\begin{aligned} & 3m > 0;m+1 > 0 \\\\ & 3m < 0;m+1 < 0 \\\\ \\end{aligned} \\right.$$\\Leftrightarrow \\left[ \\begin{aligned} & m>0;m>-1 \\\\ & m<0;m<-1 \\\\ \\end{aligned} \\right.$ $\\Leftrightarrow \\left[ \\begin{aligned} & m > 0 \\\\ & m < -1 \\\\ \\end{aligned} \\right.$ <br\/>+ $ P > 0\\Leftrightarrow \\dfrac{4m}{m+1}>0$$\\Leftrightarrow \\left[ \\begin{aligned} & 4m > 0;m+1 > 0 \\\\ & 4m < 0;m+1 < 0 \\\\ \\end{aligned} \\right.$$\\Leftrightarrow \\left[ \\begin{aligned} & m>0;m>-1 \\\\ & m<0;m<-1 \\\\ \\end{aligned} \\right.$$\\Leftrightarrow \\left[ \\begin{aligned} & m > 0 \\\\ & m < -1 \\\\ \\end{aligned} \\right.$ <br\/>K\u1ebft h\u1ee3p c\u00e1c k\u1ebft qu\u1ea3, ta c\u00f3: $ -\\dfrac{16}{7} < m < -1 $ <br\/>V\u1eady v\u1edbi $ -\\dfrac{16}{7} < m < -1 $ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m ${{x}_{2}}>{{x}_{1}} > 0$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":907},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":" <span class='basic_left'>Cho parabol (P): $y=x^2$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d): $y= 2x-2m+9.$ T\u00ecm $m$ \u0111\u1ec3 \u0111\u01b0\u1eddng th\u1eb3ng (d) c\u1eaft (P) t\u1ea1i 2 \u0111i\u1ec3m n\u1eb1m v\u1ec1 hai ph\u00eda c\u1ee7a tr\u1ee5c tung.<\/span> ","select":["A. $m<-\\dfrac{9}{2}$ ","B. $m>-\\dfrac{9}{2}$ ","C. $m<\\dfrac{9}{2}$ ","D. $m>\\dfrac{9}{2}$ "],"hint":"(d) c\u1eaft (P) t\u1ea1i 2 \u0111i\u1ec3m n\u1eb1m v\u1ec1 hai ph\u00eda c\u1ee7a tr\u1ee5c tung, n\u00ean m\u1ed9t \u0111i\u1ec3m s\u1ebd c\u00f3 ho\u00e0nh \u0111\u1ed9 \u00e2m, \u0111i\u1ec3m c\u00f2n l\u1ea1i mang ho\u00e0nh \u0111\u1ed9 d\u01b0\u01a1ng. Khi \u0111\u00f3 b\u00e0i to\u00e1n tr\u1edf th\u00e0nh t\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u.","explain":"<span class='basic_left'>X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (d) v\u00e0 (P)<br\/>$x^2 =2x\u22122m+9$<br\/>$\\Leftrightarrow x^2\u22122x+2m\u22129=0$ (1)<br\/>\u0110\u1ec3 \u0111\u01b0\u1eddng th\u1eb3ng (d) c\u1eaft (P) t\u1ea1i 2 \u0111i\u1ec3m n\u1eb1m v\u1ec1 hai ph\u00eda c\u1ee7a tr\u1ee5c tung th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m tr\u00e1i d\u1ea5u<br\/> $\\Leftrightarrow ac<0\\Leftrightarrow 2m-9<0\\Leftrightarrow m<\\dfrac{9}{2}$ <br\/>V\u1eady $m<\\dfrac{9}{2}$ th\u00ec \u0111\u01b0\u1eddng th\u1eb3ng (d) c\u1eaft (P) t\u1ea1i 2 \u0111i\u1ec3m n\u1eb1m v\u1ec1 hai ph\u00eda c\u1ee7a tr\u1ee5c tung.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":4}]}],"id_ques":908},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>Cho parabol (P): $y={{x}^{2}}$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d): $y=mx+1$ . T\u00ecm $m$ \u0111\u1ec3 (d) c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t c\u00f3 ho\u00e0nh \u0111\u1ed9 ${{x}_{1}};{{x}_{2}}$ th\u1ecfa m\u00e3n: $x_{1}^{2}{{x}_{2}}+x_{2}^{2}{{x}_{1}}-{{x}_{1}}{{x}_{2}}=3$ <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m =$ _input_ <\/span>","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai<br\/>B\u01b0\u1edbc 2: T\u00ecm $m$ \u0111\u1ec3 (d) c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t, t\u1ee9c ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t<br\/>B\u01b0\u1edbc 3: Vi\u1ebft h\u1ec7 th\u1ee9c Vi-\u00e9t v\u00e0 bi\u1ebfn di\u1ec5n h\u1ec7 th\u1ee9c \u0111\u00e3 cho v\u1ec1 d\u1ea1ng ch\u1ec9 ch\u1ee9a t\u1ed5ng v\u00e0 t\u00edch hai nghi\u1ec7m. T\u1eeb \u0111\u00f3 t\u00ecm $m$<br\/>B\u01b0\u1edbc 4: So s\u00e1nh $m$ v\u1edbi \u0111i\u1ec1u ki\u1ec7n \u1edf b\u01b0\u1edbc 2 v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) l\u00e0<br\/> $\\begin{align} & {{x}^{2}}=mx+1 \\\\ & \\Leftrightarrow {{x}^{2}}-mx-1=0\\,\\,\\,\\,\\left( 1 \\right) \\\\ \\end{align}$ <br\/>Ta c\u00f3: $\\Delta '={{m}^{2}}+4>0$ v\u1edbi m\u1ecdi $m$<br\/> Suy ra ph\u01b0\u01a1ng tr\u00ecnh (1) lu\u00f4n c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t.<br\/>Suy ra (d) lu\u00f4n c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t<br\/> \u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=m \\\\ & {{x}_{1}}{{x}_{2}}=-1 \\\\ \\end{align} \\right.$ <br\/>Theo gi\u1ea3 thi\u1ebft, ta c\u00f3: $x_{1}^{2}{{x}_{2}}+x_{2}^{2}{{x}_{1}}-{{x}_{1}}{{x}_{2}}=3$<br\/>$\\begin{align} & \\Leftrightarrow {{x}_{1}}{{x}_{2}}\\left( {{x}_{1}}+{{x}_{2}} \\right)-{{x}_{1}}{{x}_{2}}=3 \\\\ & \\Leftrightarrow -1.m+1=3 \\\\ & \\Leftrightarrow m=-2 \\\\ \\end{align}$ <br\/>Ta c\u00f3 $m=-2$ th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n b\u00e0i to\u00e1n<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2$<\/span>"}]}],"id_ques":909},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $-\\dfrac{1}{2}$","B. $-\\dfrac{1}{3}$","C. $-\\dfrac{1}{4}$"],"ques":"<span class='basic_left'>Cho parabol (P): $y=\\dfrac{1}{2}{{x}^{2}}$ v\u00e0 \u0111\u01b0\u1eddng th\u1eb3ng (d): $y=mx-\\dfrac{1}{2}{{m}^{2}}+m+1.$ T\u00ecm $m$ \u0111\u1ec3 (d) c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t c\u00f3 ho\u00e0nh \u0111\u1ed9 ${{x}_{1}};{{x}_{2}}$ sao cho $\\left| {{x}_{1}}-{{x}_{2}} \\right|=2$","hint":"","explain":"<span class='basic_left'> Ph\u01b0\u01a1ng tr\u00ecnh ho\u00e0nh \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a (P) v\u00e0 (d) l\u00e0:<br\/> $\\begin{align} & \\dfrac{1}{2}{{x}^{2}}=mx-\\dfrac{1}{2}{{m}^{2}}+m+1 \\\\ & \\Leftrightarrow {{x}^{2}}-2mx+{{m}^{2}}-2m-2=0\\,\\,\\,\\,\\left( 1 \\right) \\\\ \\end{align}$<br\/> Ta c\u00f3: $\\Delta '={{m}^{2}}-\\left( {{m}^{2}}-2m-2 \\right)=2m+2$<br\/> \u0110\u1ec3 (d) c\u1eaft (P) t\u1ea1i hai \u0111i\u1ec3m ph\u00e2n bi\u1ec7t th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t <br\/>$\\Leftrightarrow \\Delta '>0\\Leftrightarrow 2m+2>0\\Leftrightarrow m>-1$ (*)<br\/>\u00c1p d\u1ee5ng h\u1ec7 th\u1ee9c Vi-\u00e9t, ta c\u00f3: $\\left\\{ \\begin{align} & {{x}_{1}}+{{x}_{2}}=2m \\\\ & {{x}_{1}}{{x}_{2}}={{m}^{2}}-2m-2 \\\\ \\end{align} \\right.$<br\/> Theo gi\u1ea3 thi\u1ebft, ta c\u00f3:<br\/> $\\left| {{x}_{1}}-{{x}_{2}} \\right|=2$<br\/>$\\begin{align} & \\Leftrightarrow {{\\left( {{x}_{1}}-{{x}_{2}} \\right)}^{2}}=4 \\\\ & \\Leftrightarrow {{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-4{{x}_{1}}{{x}_{2}}=4 \\\\ & \\Leftrightarrow 4{{m}^{2}}-4\\left( {{m}^{2}}-2m-2 \\right)=4 \\\\ & \\Leftrightarrow 8m=-4 \\\\ & \\Leftrightarrow m=-\\dfrac{1}{2} \\,\\text{(th\u1ecfa m\u00e3n (*))} \\\\ \\end{align}$<br\/> V\u1eady $m=-\\dfrac{1}{2}$ th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n b\u00e0i to\u00e1n."}]}],"id_ques":910}],"lesson":{"save":0,"level":2}}