{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Ch\u1ec9 c\u00f3 m\u1ed9t c\u1eb7p s\u1ed1 duy nh\u1ea5t $(x;y)$ th\u1ecfa m\u00e3n ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-4x+y-6\\sqrt{y}+13=0,$ <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?","select":["A. \u0110\u00fang ","B. Sai"],"hint":"X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u1edbi $x$ l\u00e0 \u1ea9n v\u00e0 $y$ l\u00e0 tham s\u1ed1 ho\u1eb7c \u0111\u01b0a v\u1ec1 t\u1ed5ng b\u00ecnh ph\u01b0\u01a1ng.","explain":"<span class='basic_left'>Coi $x$ l\u00e0 \u1ea9n v\u00e0 $y$ l\u00e0 tham s\u1ed1 $\\left( y\\ge 0 \\right).$ <br\/>Khi \u0111\u00f3 ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u1edbi c\u00e1c h\u1ec7 s\u1ed1 $a=1;b'=-2;c=y-6\\sqrt{y}+13$ <br\/>Ta c\u00f3 $\\Delta '=4-\\left( y-6\\sqrt{y}+13 \\right)=-\\left( y-6\\sqrt{y}+9 \\right)=-{{\\left( \\sqrt{y}-3 \\right)}^{2}} \\le 0$ v\u1edbi m\u1ecdi $ y\\ge 0 $ <br\/>V\u00ec $\\Delta '\\le 0$ n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m khi $\\Delta '=0\\Leftrightarrow \\sqrt{y}-3=0\\Leftrightarrow y=9$ <br\/>Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p $x=-\\dfrac{b'}{a}=2$ <br\/>V\u1eady c\u1eb7p s\u1ed1 $(2;9)$ l\u00e0 c\u1eb7p s\u1ed1 duy nh\u1ea5t th\u1ecfa m\u00e3n ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho.<br\/>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0111\u00fang<br\/><span class='basic_pink'> V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i trong b\u00e0i to\u00e1n tr\u00ean \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 <b>''\u0110\u1eb7t tham s\u1ed1 m\u1edbi\u2019\u2019:<\/b> T\u1eeb m\u1ed9t ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai \u1ea9n s\u1ed1, ta xem m\u1ed9t \u1ea9n l\u00e0 tham s\u1ed1 r\u1ed3i gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea5y theo \u1ea9n c\u00f2n l\u1ea1i<br\/><b>C\u00e1ch gi\u1ea3i kh\u00e1c:<\/b> \u0110\u01b0a v\u1ec1 t\u1ed5ng b\u00ecnh ph\u01b0\u01a1ng<br\/>$\\begin{aligned} & {{x}^{2}}-4x+y-6\\sqrt{y}+13=0 \\\\ & \\Leftrightarrow \\left( {{x}^{2}}-4x+4 \\right)+\\left( y-6\\sqrt{y}+9 \\right)=0 \\\\ & \\Leftrightarrow {{\\left( x-2 \\right)}^{2}}+{{\\left( \\sqrt{y}-3 \\right)}^{2}}=0 \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & x-2=0 \\\\ & \\sqrt{y}-3=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x=2 \\\\ & y=9 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<\/span>","column":2}]}],"id_ques":861},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh ${{b}^{2}}{{x}^{2}}-\\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \\right)x+{{c}^{2}}=0$ v\u1edbi $a,$ $b,$ $c$ l\u00e0 \u0111\u1ed9 d\u00e0i ba c\u1ea1nh trong tam gi\u00e1c <br\/>Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang:<\/span>","select":["<span class='basic_left'>A. Ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t <\/span>","<span class='basic_left'>B. Ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 nghi\u1ec7m k\u00e9p<\/span>","<span class='basic_left'>C. Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<\/span>"],"hint":"Ph\u00e2n t\u00edch $\\Delta$ th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i \u00e1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c \u0111\u1ec3 x\u00e9t d\u1ea5u $\\Delta.$","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{b}^{2}}{{x}^{2}}-\\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \\right)x+{{c}^{2}}=0$ c\u00f3<br\/>$\\begin{align} & \\Delta =\\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \\right)^{2}-4{{b}^{2}}{{c}^{2}} \\\\ & \\,\\,\\,\\,\\,=\\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}}-2bc \\right)\\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}}+2bc \\right) \\\\ & \\,\\,\\,\\,\\,=\\left[ {{\\left( b-c \\right)}^{2}}-{{a}^{2}} \\right].\\left[ {{\\left( b+c \\right)}^{2}}-{{a}^{2}} \\right] \\\\ & \\,\\,\\,\\,\\,=\\left( b-c-a \\right)\\left( b-c+a \\right)\\left( b+c-a \\right)\\left( b+c+a \\right) \\\\ \\end{align}$<br\/>V\u00ec $a, b, c$ l\u00e0 ba c\u1ea1nh trong tam gi\u00e1c n\u00ean $a+b+c>0$ v\u00e0 $\\left\\{ \\begin{align} & a+b>c \\\\ & a+c>b \\\\ & b+c>a \\\\ \\end{align} \\right.$ (b\u1ea5t \u0111\u1eb3ng th\u1ee9c tam gi\u00e1c)<br\/>Suy ra $\\left\\{ \\begin{align} & b-c+a>0 \\\\ & b-c-a<0 \\\\ & b+c-a>0 \\\\ & a+b+c>0 \\\\ \\end{align} \\right.$ <br\/>Do \u0111\u00f3 $\\Delta =\\left( b-c-a \\right)\\left( b-c+a \\right)\\left( b+c-a \\right)\\left( b+c+a \\right)<0$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho v\u00f4 nghi\u1ec7m<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":1}]}],"id_ques":862},{"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":10,"width":40,"content":"","type_input":"","ques":"Cho hai ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{2}}+x+a=0$ v\u00e0 ${{x}^{2}}+ax+1=0$ v\u1edbi $a$ l\u00e0 tham s\u1ed1.<br\/> X\u00e1c \u0111\u1ecbnh $a$ \u0111\u1ec3 hai ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean c\u00f3 nghi\u1ec7m chung<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> $a=$_input_ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/><b>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i<\/b><br\/>B\u01b0\u1edbc 1: Gi\u1ea3 s\u1eed ${{x}_{0}}$ l\u00e0 nghi\u1ec7m chung c\u1ee7a hai ph\u01b0\u01a1ng tr\u00ecnh. Thay $x={{x}_{0}}$ v\u00e0o hai ph\u01b0\u01a1ng tr\u00ecnh ta \u0111\u01b0\u1ee3c h\u1ec7 v\u1edbi c\u00e1c \u1ea9n l\u00e0 tham s\u1ed1.<br\/>B\u01b0\u1edbc 2: Gi\u1ea3i h\u1ec7 t\u00ecm tham s\u1ed1<br\/>B\u01b0\u1edbc 3: Th\u1eed l\u1ea1i v\u1edbi tham s\u1ed1 v\u1eeba t\u00ecm, hai ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m chung hay kh\u00f4ng.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Gi\u1ea3 s\u1eed ${{x}_{0}}$ l\u00e0 nghi\u1ec7m chung c\u1ee7a hai ph\u01b0\u01a1ng tr\u00ecnh. Ta c\u00f3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\left\\{ \\begin{align} & x_{0}^{2}+{{x}_{0}}+a=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & x_{0}^{2}+a{{x}_{0}}+1=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align} \\right.$ <br\/>L\u1ea5y t\u1eebng v\u1ebf c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1eeb t\u1eebng v\u1ebf c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh (2), ta \u0111\u01b0\u1ee3c:<br\/>${{x}_{0}}\\left( 1-a \\right)+a-1=0\\Leftrightarrow \\left( 1-a \\right)\\left( {{x}_{0}}-1 \\right)=0\\Leftrightarrow \\left[ \\begin{align} & a=1 \\\\ & {{x}_{0}}=1 \\\\ \\end{align} \\right.$ <br\/>+ V\u1edbi $a=1$ , ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+x+1=0$ v\u00f4 nghi\u1ec7m.<br\/>+ V\u1edbi ${{x}_{0}}=1$ thay v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh (1), ta c\u00f3: $2+a=0\\Leftrightarrow a=-2$ <br\/>Ng\u01b0\u1ee3c l\u1ea1i v\u1edbi $a=-2,$ ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+x-2=0$ c\u00f3 nghi\u1ec7m l\u00e0 ${{x}_{1}}=1;{{x}_{2}}=-2$ v\u00e0 ph\u01b0\u01a1ng tr\u00ecnh ${{x}_{2}}-2x+1=0$ c\u00f3 nghi\u1ec7m k\u00e9p l\u00e0 $x=1$<br\/>V\u1eady v\u1edbi $a=-2$ th\u00ec hai ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m chung l\u00e0 $x=1$<br\/><span class='basic_green'>C\u00e1ch 2:<\/span><br\/>D\u1ec5 th\u1ea5y $x=0$ kh\u00f4ng l\u00e0 nghi\u1ec7m chung c\u1ee7a hai ph\u01b0\u01a1ng tr\u00ecnh n\u00ean $x \\ne 0.$<br\/>T\u1eeb hai ph\u01b0\u01a1ng tr\u00ecnh ta bi\u1ec3u th\u1ecb $a$ qua $x,$ ta c\u00f3:<br\/>$\\begin{align} & a=-{{x}^{2}}-x; \\\\ & a=\\dfrac{-{{x}^{2}}-1}{x} \\\\ \\end{align}$ <br\/>Suy ra $\\begin{align} & \\dfrac{-{{x}^{2}}-1}{x}=-{{x}^{2}}-x\\Leftrightarrow {{x}^{2}}+1={{x}^{3}}+{{x}^{2}} \\\\ & \\Leftrightarrow {{x}^{3}}=1\\Leftrightarrow x=1 \\\\ \\end{align}$<br\/> Khi \u0111\u00f3 $a=-1-1=-2$<br\/>Ki\u1ec3m tra v\u1edbi $a=-2,$ hai ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m chung kh\u00f4ng v\u00e0 k\u1ebft lu\u1eadn.<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2$ <\/span><\/span>"}]}],"id_ques":863},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"Cho s\u1ed1 d\u01b0\u01a1ng $a,$ $b$ th\u1ecfa m\u00e3n $a+b=4\\sqrt{ab}.$ T\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{a}{b}$ ","select":["A. $\\dfrac{a}{b}=2\\pm \\sqrt{3}$ ","B. $\\dfrac{a}{b}=4\\pm 2\\sqrt{3}$","C. $\\dfrac{a}{b}=7\\pm 4\\sqrt{3}$","D. $\\dfrac{a}{b}=-7\\pm 4\\sqrt{3}$"],"hint":"Chia c\u1ea3 hai v\u1ebf c\u1ee7a $a+b=4\\sqrt{ab}$ cho $b,$ r\u1ed3i d\u00f9ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t \u1ea9n ph\u1ee5 \u0111\u1ec3 gi\u1ea3i","explain":"<span class='basic_left'>V\u00ec $b>0$ n\u00ean chia c\u1ea3 hai v\u1ebf c\u1ee7a $a+b=4\\sqrt{ab}$ cho $b,$ ta \u0111\u01b0\u1ee3c:<br\/>$\\dfrac{a}{b}+1=4\\sqrt{\\dfrac{a}{b}}$ <br\/>\u0110\u1eb7t $t=\\sqrt{\\dfrac{a}{b}}\\,\\,\\,,t>0$ . Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>${{t}^{2}}+1=4t\\Leftrightarrow {{t}^{2}}-4t+1=0$ <br\/>$\\Delta '=4-1=3$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $t=2\\pm \\sqrt{3}$ (th\u1ecfa m\u00e3n)<br\/>Suy ra $\\dfrac{a}{b}={{t}^{2}}={{\\left( 2\\pm \\sqrt{3} \\right)}^{2}}=7\\pm 4\\sqrt{3}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":864},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"T\u00ecm $a$ \u0111\u1ec3 h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{aligned} & x-y=3\\,\\,\\,\\,\\,\\, \\\\ & {{x}^{2}}+{{y}^{2}}+2y=a \\\\ \\end{aligned} \\right.$ c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t","select":["A. $a >-1$ ","B. $a>1$","C. $a<1$","D. $a<-1$"],"hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p th\u1ebf","explain":"<span class='basic_left'>T\u1eeb ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 nh\u1ea5t, bi\u1ec3u di\u1ec5n $x$ theo $y$ r\u1ed3i th\u1ebf v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh th\u1ee9 hai, ta c\u00f3: <br\/> $\\begin{aligned} & \\left\\{ \\begin{aligned} & x=y+3 \\\\ & {{\\left( y+3 \\right)}^{2}}+{{y}^{2}}+2y=a \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & x=y+3 \\\\ & 2{{y}^{2}}+8y+9-a=0\\,\\,\\left( * \\right) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/> X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 $\\Delta '=16-2\\left( 9-a \\right)=2a-2$ <br\/>H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t $\\Leftrightarrow $ ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t<br\/>$\\Leftrightarrow \\Delta '>0\\Leftrightarrow 2a-2>0\\Leftrightarrow a>1$<br\/> V\u1eady $a>1$ th\u00ec h\u1ec7 \u0111\u00e3 cho c\u00f3 2 nghi\u1ec7m ph\u00e2n bi\u1ec7t. <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><\/span>","column":4}]}],"id_ques":865},{"time":24,"part":[{"time":3,"title":"Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+mx+3=0$ v\u1edbi $m$ nguy\u00ean. T\u00ecm nghi\u1ec7m h\u1eefu t\u1ec9 c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh","title_trans":"H\u00e3y s\u1eafp x\u1ebfp c\u00e1c c\u00e2u sau \u0111\u1ec3 \u0111\u01b0\u1ee3c l\u1eddi gi\u1ea3i \u0111\u00fang","temp":"sequence","correct":[[[3],[5],[2],[6],[1],[4]]],"list":[{"point":10,"image":"img\/1.png","left":["Suy ra $m+k$ v\u00e0 $m-k$ l\u00e0 \u01b0\u1edbc c\u1ee7a $12.$ M\u00e0 $\\left( m+k \\right)-\\left( m-k \\right)=2k$ l\u00e0 s\u1ed1 ch\u1eb5n n\u00ean $m+k$ v\u00e0 $m-k$ c\u00f9ng ch\u1eb5n ho\u1eb7c c\u00f9ng l\u1ebb. Do t\u00edch c\u1ee7a ch\u00fang l\u00e0 s\u1ed1 ch\u1eb5n (12) n\u00ean ch\u00fang c\u00f9ng ch\u1eb5n","Gi\u1ea3i h\u1ec7 (I), ta \u0111\u01b0\u1ee3c $m=4,$ $k=2.$ Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (*) l\u00e0 $x^2+4x+3=0$ c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=-1;{{x}_{2}}=-3$<br\/>Gi\u1ea3i h\u1ec7 (II), ta \u0111\u01b0\u1ee3c $m=-4,$ $k=2.$ Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (*) l\u00e0 $x^2-4x+3=0$ c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=1;{{x}_{2}}=3$ ","Suy ra ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m h\u1eefu t\u1ec9 $\\Leftrightarrow \\Delta $ l\u00e0 s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng, t\u1ee9c l\u00e0 <br\/>${{m}^{2}}-12={{k}^{2}}\\,\\,\\,\\left( k\\in \\mathbb{N} \\right)\\Leftrightarrow {{m}^{2}}-{{k}^{2}}=12\\Leftrightarrow \\left( m+k \\right)\\left( m-k \\right)=12$","K\u1ebft lu\u1eadn: <br\/>V\u1edbi $m=4,$ ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m h\u1eefu t\u1ec9 l\u00e0 $-1$ v\u00e0 $-3$<br\/>V\u1edbi $m=-4,$ ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m h\u1eefu t\u1ec9 l\u00e0 $1$ v\u00e0 $3$","Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+mx+3=0$(*) c\u00f3 $\\Delta ={{m}^{2}}-12 \\in \\mathbb{Z}$ v\u1edbi m\u1ecdi $m \\in \\mathbb{Z}$ ","L\u1ea1i c\u00f3 $m+k\\ge m-k$ n\u00ean ta c\u00f3 hai tr\u01b0\u1eddng h\u1ee3p:<br\/>$\\left\\{ \\begin{align} & m+k=6 \\\\ & m-k=2 \\\\ \\end{align} \\right.\\,\\,\\,\\left( I \\right)$ v\u00e0 $\\left\\{ \\begin{align} & m+k=-2 \\\\ & m-k=-6 \\\\ \\end{align} \\right.$ (II)"],"top":120,"hint":"Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 $\\Delta$ lu\u00f4n nh\u1eadn gi\u00e1 tr\u1ecb nguy\u00ean v\u1edbi m\u1ecdi $m$ nguy\u00ean n\u00ean ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m h\u1eefu t\u1ec9 $\\Leftrightarrow\\Delta$ l\u00e0 s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng.","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+mx+3=0$(*) c\u00f3 $\\Delta ={{m}^{2}}-12 \\in \\mathbb{Z}$ v\u1edbi m\u1ecdi $m \\in \\mathbb{Z}$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m h\u1eefu t\u1ec9 $\\Leftrightarrow \\Delta $ l\u00e0 s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng, t\u1ee9c l\u00e0 <br\/>${{m}^{2}}-12={{k}^{2}}\\,\\,\\,\\left( k\\in \\mathbb{N} \\right)\\Leftrightarrow {{m}^{2}}-{{k}^{2}}=12\\Leftrightarrow \\left( m+k \\right)\\left( m-k \\right)=12$ <br\/>Suy ra $m+k$ v\u00e0 $m-k$ l\u00e0 \u01b0\u1edbc c\u1ee7a $12.$ <br\/>M\u00e0 $\\left( m+k \\right)-\\left( m-k \\right)=2k$ l\u00e0 s\u1ed1 ch\u1eb5n n\u00ean $m+k$ v\u00e0 $m-k$ c\u00f9ng ch\u1eb5n ho\u1eb7c c\u00f9ng l\u1ebb.<br\/> Do t\u00edch c\u1ee7a ch\u00fang l\u00e0 s\u1ed1 ch\u1eb5n $(12)$ n\u00ean ch\u00fang c\u00f9ng ch\u1eb5n<br\/>L\u1ea1i c\u00f3 $m+k\\ge m-k$ n\u00ean ta c\u00f3 hai tr\u01b0\u1eddng h\u1ee3p:<br\/>$\\left\\{ \\begin{align} & m+k=6 \\\\ & m-k=2 \\\\ \\end{align} \\right.\\,\\,\\,\\left( I \\right)$ v\u00e0 $\\left\\{ \\begin{align} & m+k=-2 \\\\ & m-k=-6 \\\\ \\end{align} \\right.$ (II)<br\/>Gi\u1ea3i h\u1ec7 (I), ta \u0111\u01b0\u1ee3c $m=4,$ $k=2.$ Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (*) l\u00e0 $x^2+4x+3=0$ c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=-1;{{x}_{2}}=-3$ <br\/>Gi\u1ea3i h\u1ec7 (II), ta \u0111\u01b0\u1ee3c $m=-4,$ $k=2.$ Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (*) l\u00e0 $x^2-4x+3=0$ c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=1;{{x}_{2}}=3$ <br\/>K\u1ebft lu\u1eadn: <br\/>V\u1edbi $m=4,$ ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m h\u1eefu t\u1ec9 l\u00e0 $-1$ v\u00e0 $-3$<br\/>V\u1edbi $m=-4,$ ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m h\u1eefu t\u1ec9 l\u00e0 $1$ v\u00e0 $3.$<\/span>"}]}],"id_ques":866},{"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":[[["3"],["-3"],["0"]]],"list":[{"point":10,"width":40,"content":"","type_input":"","input_hint":["frac"],"ques":"T\u00ecm c\u00e1c s\u1ed1 nguy\u00ean $n$ \u0111\u1ec3 c\u00e1c nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh sau l\u00e0 c\u00e1c s\u1ed1 nguy\u00ean:<br\/>${{x}^{2}}-\\left( 4+n \\right)x+2n=0$ <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $n \\in$ {_input_;_input_;_input_}","hint":"Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 nghi\u1ec7m nguy\u00ean n\u1ebfu $\\Delta $ l\u00e0 s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>B\u01b0\u1edbc 1: T\u00ednh $\\Delta $ v\u00e0 l\u1eadp lu\u1eadn bi\u1ec3u di\u1ec5n $\\Delta $ d\u01b0\u1edbi d\u1ea1ng s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng.<br\/>B\u01b0\u1edbc 2: \u0110\u01b0a \u0111\u1eb3ng th\u1ee9c tr\u00ean v\u1ec1 d\u1ea1ng $(a-b)(a+b)=2k.$ Khi \u0111\u00f3 $a-b;a+b$ l\u00e0 \u01b0\u1edbc c\u1ee7a $2k.$<br\/> Ta \u0111\u00e1nh gi\u00e1 $a-b$ v\u00e0 $a+b$ l\u00e0 c\u00e1c s\u1ed1 ch\u1eb5n, t\u1eeb \u0111\u00f3 t\u00ecm $n.$<br\/>B\u01b0\u1edbc 3: Th\u1eed l\u1ea1i v\u1edbi c\u00e1c gi\u00e1 tr\u1ecb $n$ v\u1eeba t\u00ecm \u0111\u01b0\u1ee3c th\u00ec ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m nguy\u00ean kh\u00f4ng v\u00e0 k\u1ebft lu\u1eadn b\u00e0i to\u00e1n. <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-\\left( 4+n \\right)x+2n=0$ (1) c\u00f3 $\\Delta ={{\\left( 4+n \\right)}^{2}}-8n={{n}^{2}}+16>0$ <br\/>Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t<br\/>Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 s\u1ed1 nguy\u00ean n\u1ebfu $\\Delta $ l\u00e0 s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng, t\u1ee9c ${{n}^{2}}+16={{k}^{2}},\\,\\,k\\in \\mathbb{N}$ <br\/>Ta c\u00f3 ${{n}^{2}}-{{k}^{2}}=-16\\Leftrightarrow \\left( n-k \\right)\\left( n+k \\right)=-16$ <br\/>Do $(n+k)-(n-k)=2k$ n\u00ean $n+k$ v\u00e0 $n-k$ c\u00f9ng ch\u1eb5n ho\u1eb7c c\u00f9ng l\u1ebb. M\u00e0 t\u00edch c\u1ee7a ch\u00fang l\u00e0 s\u1ed1 ch\u1eb5n n\u00ean $n+k$ v\u00e0 $n-k$ c\u00f9ng ch\u1eb5n.<br\/> L\u1ea1i c\u00f3 $n+k\\ge n-k$; $n+k$ v\u00e0 $n-k$ l\u00e0 \u01b0\u1edbc c\u1ee7a $16$ n\u00ean ta c\u00f3 b\u1ea3ng sau:<br\/><br\/><table><tr><td>$n+k$<\/td><td>$8$<\/td><td>$2$<\/td><td>$4$<\/td><\/tr><tr><td>$n-k$<\/td><td>$-2$<\/td><td>$-8$<\/td><td>$-4$<\/td><\/tr><tr><td>$n$<\/td><td>$3$<\/td><td>$-3$<\/td><td>$0$<\/td><\/tr><\/table><br\/>V\u1edbi $n=3$ th\u00ec (1) tr\u1edf th\u00e0nh $x^2-7x+6=0.$ Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $1$ v\u00e0 $6$<br\/>V\u1edbi $n=-3$ th\u00ec (1) tr\u1edf th\u00e0nh $x^2-x-6=0.$ Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 $-2$ v\u00e0 $3.$<br\/>V\u1edbi $n=0,$ th\u00ec (1) tr\u1edf th\u00e0nh $x^2-4x=0,$ n\u00f3 c\u00f3 c\u00e1c nghi\u1ec7m l\u00e0 $0$ v\u00e0 $4.$<br\/>V\u1eady v\u1edbi $n\\in \\text{ }\\!\\!\\{\\!\\!\\text{ }3;-3;0\\}$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m nguy\u00ean<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3;-3;0$ <\/span><\/span>"}]}],"id_ques":867},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":10,"ques":"Cho h\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh $\\left\\{ \\begin{align} & {{x}^{3}}+{{y}^{2}}=2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\left( 1 \\right) \\\\ & {{x}^{2}}+xy+{{y}^{2}}-y=0\\,\\,\\,\\,\\,\\,\\,\\,\\left( 2 \\right) \\\\ \\end{align} \\right.$ <br\/> H\u1ec7 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 bao nhi\u00eau nghi\u1ec7m?","select":["A. $0 $ ","B. $1$","C. $2$","D. $3$"],"hint":"Gi\u1ea3 s\u1eed h\u1ec7 \u0111\u00e3 cho c\u00f3 nghi\u1ec7m. Ta x\u00e9t ph\u01b0\u01a1ng tr\u00ecnh (2) b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p ''\u0110\u1eb7t tham s\u1ed1 m\u1edbi'': Coi 1 \u1ea9n l\u00e0 tham s\u1ed1. T\u1eeb \u0111\u00f3 \u0111\u00e1nh gi\u00e1 $x$ v\u00e0 $y$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 nghi\u1ec7m","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Gi\u1ea3 s\u1eed h\u1ec7 \u0111\u00e3 cho c\u00f3 nghi\u1ec7m.<br\/>B\u01b0\u1edbc 1: Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh (2) d\u01b0\u1edbi d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u1edbi \u1ea9n $y$ v\u00e0 $x$ l\u00e0 tham s\u1ed1. T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $x$ \u0111\u1ec3 (2) c\u00f3 nghi\u1ec7m<br\/>B\u01b0\u1edbc 2: Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh (2) d\u01b0\u1edbi d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u1edbi \u1ea9n $x$ v\u00e0 $y$ l\u00e0 tham s\u1ed1. T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $y$ \u0111\u1ec3 (2) c\u00f3 nghi\u1ec7m<br\/>B\u01b0\u1edbc 3: T\u1eeb \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $x$ v\u00e0 $y,$ ta \u0111\u00e1nh gi\u00e1 ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00e0 k\u1ebft lu\u1eadn b\u00e0i to\u00e1n<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Gi\u1ea3 s\u1eed h\u1ec7 \u0111\u00e3 cho c\u00f3 nghi\u1ec7m, khi \u0111\u00f3 (2) c\u00f3 nghi\u1ec7m v\u1edbi \u1ea9n $y$ (coi $x$ l\u00e0 tham s\u1ed1).<br\/>Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh (2) $\\Leftrightarrow {{y}^{2}}+\\left( x-1 \\right)y+{{x}^{2}}=0$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 $a=1;b=x-1;c=x^2$<br\/>$\\begin{align} & {{\\Delta }_{1}}={{\\left( x-1 \\right)}^{2}}-4{{x}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,={{x}^{2}}-2x+1-4{{x}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,=-\\left( 3{{x}^{2}}+2x-1 \\right) \\\\ & \\,\\,\\,\\,\\,\\,=-\\left( x+1 \\right)\\left( 3x-1 \\right) \\\\ \\end{align}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 nghi\u1ec7m khi v\u00e0 ch\u1ec9 khi ${{\\Delta }_{1}}\\ge 0\\Leftrightarrow \\left( x+1 \\right)\\left( 3x-1 \\right)\\le 0$<br\/>$\\Leftrightarrow \\left[ \\begin{aligned} & \\left\\{ \\begin{aligned} & x+1\\le 0 \\\\ & 3x-1\\ge 0 \\\\ \\end{aligned} \\right. \\\\ & \\left\\{ \\begin{aligned} & x+1\\ge 0 \\\\ & 3x-1\\le 0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & \\dfrac{1}{3}\\le x\\le -1 \\\\ & -1\\le x\\le \\dfrac{1}{3} \\\\ \\end{aligned} \\right.$ $\\Leftrightarrow -1\\le x\\le \\dfrac{1}{3}$ (*)<br\/>Vi\u1ebft ph\u01b0\u01a1ng tr\u00ecnh (2) theo \u1ea9n $x$ (coi $y$ l\u00e0 tham s\u1ed1), ta c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 $a=1;b=y;c={{y}^{2}}-y$<br\/>Suy ra ${{\\Delta }_{2}}={{y}^{2}}-4\\left( {{y}^{2}}-y \\right)=-3{{y}^{2}}+4y=-y\\left( 3y-4 \\right)$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh (2) c\u00f3 nghi\u1ec7m khi v\u00e0 ch\u1ec9 khi ${{\\Delta }_{2}}\\ge 0\\Leftrightarrow y\\left( 3y-4 \\right)\\le 0$<br\/>$\\Rightarrow 3y-4 \\le 0$ (v\u00ec $y \\ge 0$) $\\Leftrightarrow 0\\le y\\le \\dfrac{4}{3}$(**)<br\/>T\u1eeb (*) v\u00e0 (**) suy ra ${{x}^{3}}+{{y}^{2}}\\le {{\\left( \\dfrac{1}{3} \\right)}^{3}}+{{\\left( \\dfrac{4}{3} \\right)}^{2}}=\\dfrac{49}{27}<2$ <br\/>Suy ra (1) v\u00f4 nghi\u1ec7m.<br\/>V\u1eady h\u1ec7 \u0111\u00e3 cho v\u00f4 nghi\u1ec7m<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 A <\/span><\/span>","column":4}]}],"id_ques":868},{"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":10,"width":40,"content":"","type_input":"","ques":"T\u00ecm s\u1ed1 nguy\u00ean t\u1ed1 $p$ bi\u1ebft r\u1eb1ng ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+px-12p=0$ c\u00f3 hai nghi\u1ec7m \u0111\u1ec1u l\u00e0 nh\u1eefng s\u1ed1 nguy\u00ean <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $p=$_input_ ","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>M\u1ed9t s\u1ed1 ki\u1ebfn th\u1ee9c c\u1ea7n s\u1eed d\u1ee5ng:<br\/>+ N\u1ebfu ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 nghi\u1ec7m nguy\u00ean th\u00ec $\\Delta $ l\u00e0 s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng<br\/>+ S\u1ed1 nguy\u00ean t\u1ed1 ch\u1ec9 c\u00f3 \u01b0\u1edbc l\u00e0 $1$ v\u00e0 ch\u00ednh n\u00f3.<br\/>+ Khi ph\u00e2n t\u00edch 1 s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng ra th\u1eeba s\u1ed1 nguy\u00ean t\u1ed1, ta \u0111\u01b0\u1ee3c c\u00e1c th\u1eeba s\u1ed1 l\u00e0 l\u0169y th\u1eeba c\u1ee7a s\u1ed1 nguy\u00ean t\u1ed1 v\u1edbi s\u1ed1 m\u0169 ch\u1eb5n.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+px-12p=0$ c\u00f3 $\\Delta ={{p}^{2}}+48p$ <br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 nghi\u1ec7m nguy\u00ean th\u00ec $\\Delta $ l\u00e0 s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng<br\/>$\\Leftrightarrow p\\left( p+48 \\right)$ l\u00e0 s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng<br\/>$\\Rightarrow p+48$ chia h\u1ebft cho $p$<br\/>$\\Rightarrow 48$ chia h\u1ebft cho $p$<br\/>\u01af\u1edbc nguy\u00ean t\u1ed1 c\u1ee7a $48$ l\u00e0 $2$ v\u00e0 $3$ n\u00ean $p\\in \\left\\{ 2;3 \\right\\}$ <br\/>N\u1ebfu $p=2$ th\u00ec $\\Delta =100={{10}^{2}}$ . Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh $x^2+2x-24=0$ c\u00f3 hai nghi\u1ec7m l\u00e0 $4$ v\u00e0 $-6$<br\/>N\u1ebfu $p=3$ th\u00ec $\\Delta =153$ kh\u00f4ng l\u00e0 s\u1ed1 ch\u00ednh ph\u01b0\u01a1ng n\u00ean lo\u1ea1i<br\/>V\u1eady $p=2$ th\u00ec th\u1ecfa m\u00e3n y\u00eau c\u00e2u 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":869},{"time":24,"part":[{"time":3,"title":"Cho $ac\\ge 2\\left( b+d \\right)$ . Ch\u1ee9ng minh r\u1eb1ng c\u00f3 \u00edt nh\u1ea5t m\u1ed9t trong hai ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+ax+b=0$ v\u00e0 ${{x}^{2}}+cx+d=0$ c\u00f3 nghi\u1ec7m","title_trans":"H\u00e3y s\u1eafp x\u1ebfp c\u00e1c c\u00e2u sau \u0111\u1ec3 \u0111\u01b0\u1ee3c l\u1eddi gi\u1ea3i \u0111\u00fang","temp":"sequence","correct":[[[3],[1],[5],[4],[2]]],"list":[{"point":10,"image":"img\/1.png","left":["Theo gi\u1ea3 thi\u1ebft: $ac\\ge 2\\left( b+d \\right)$<br\/>Suy ra ${{\\Delta }_{1}}+{{\\Delta }_{2}}\\ge {{a}^{2}}+{{c}^{2}}-2ac={{\\left( a-c \\right)}^{2}}\\ge 0.$<br\/> V\u1eady ${{\\Delta }_{1}}+{{\\Delta }_{2}} \\ge 0$"," Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+ax+b=0$ c\u00f3 ${{\\Delta }_{1}}={{a}^{2}}-4b;$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+cx+d=0$ c\u00f3 ${{\\Delta }_{2}}={{c}^{2}}-4d$","V\u1eady t\u1ed3n t\u1ea1i \u00edt nh\u1ea5t 1 bi\u1ec3u th\u1ee9c ${{\\Delta }_{1}}$ ho\u1eb7c ${{\\Delta }_{2}}$ kh\u00f4ng \u00e2m. Do \u0111\u00f3 t\u1ed3n t\u1ea1i \u00edt nh\u1ea5t m\u1ed9t trong hai ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m.","Gi\u1ea3 s\u1eed ${{\\Delta }_{1}}<0;{{\\Delta }_{2}}<0$ . Khi \u0111\u00f3 ${{\\Delta }_{1}}+{{\\Delta }_{2}}<0$ (m\u00e2u thu\u1eabn v\u1edbi ch\u1ee9ng minh tr\u00ean).","Ta c\u00f3: ${{\\Delta }_{1}}+{{\\Delta }_{2}}={{a}^{2}}+{{c}^{2}}-4\\left( b+d \\right)$ "],"top":105,"hint":"Ch\u1ee9ng minh ${{\\Delta }_{1}}+{{\\Delta }_{2}}\\ge 0$ ","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+ax+b=0$ c\u00f3 ${{\\Delta }_{1}}={{a}^{2}}-4b;$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+cx+d=0$ c\u00f3 ${{\\Delta }_{2}}={{c}^{2}}-4d$<br\/>Suy ra ${{\\Delta }_{1}}+{{\\Delta }_{2}}={{a}^{2}}+{{c}^{2}}-4\\left( b+d \\right)$ <br\/>Theo gi\u1ea3 thi\u1ebft: $ac\\ge 2\\left( b+d \\right)$<br\/>Suy ra ${{\\Delta }_{1}}+{{\\Delta }_{2}}={{a}^{2}}+{{c}^{2}}-4\\left( b+d \\right) \\ge {{a}^{2}}+{{c}^{2}}-2ac={{\\left( a-c \\right)}^{2}}\\ge 0$ <br\/> V\u1eady ${{\\Delta }_{1}}+{{\\Delta }_{2}} \\ge 0$<br\/>Gi\u1ea3 s\u1eed ${{\\Delta }_{1}}<0;{{\\Delta }_{2}}<0$ . Khi \u0111\u00f3 ${{\\Delta }_{1}}+{{\\Delta }_{2}}<0$ (m\u00e2u thu\u1eabn v\u1edbi ch\u1ee9ng minh tr\u00ean).<br\/>V\u1eady t\u1ed3n t\u1ea1i \u00edt nh\u1ea5t 1 bi\u1ec3u th\u1ee9c ${{\\Delta }_{1}}$ ho\u1eb7c ${{\\Delta }_{2}}$ kh\u00f4ng \u00e2m.<br\/> Do \u0111\u00f3 t\u1ed3n t\u1ea1i \u00edt nh\u1ea5t m\u1ed9t trong hai ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m.<\/span>"}]}],"id_ques":870}],"lesson":{"save":0,"level":3}}