{"segment":[{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng trong b\u1ea3ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["50"],["-1000"],["-2"],["0"],["1"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 $a,$ $b,$ $c$ c\u1ee7a c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai:<\/span><br\/><table><tr><th>Ph\u01b0\u01a1ng tr\u00ecnh<br><\/th><th>$a$<br><\/th><th>$b$<br><\/th><th>$c$<br><\/th><\/tr><tr><td>${{x}^{2}}+50x-1000=0$ <\/td><td>_input_<\/td><td>_input_<\/td><td>_input_<\/td><\/tr><tr><td>$1-2{{x}^{2}}=0$ <\/td><td>_input_<\/td><td>_input_<\/td><td>_input_<\/td><\/tr><\/table>","hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai m\u1ed9t \u1ea9n l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 d\u1ea1ng $a{{x}^{2}}+bx+c=0$ trong \u0111\u00f3 $x$ l\u00e0 \u1ea9n; $a,b,c$ l\u00e0 c\u00e1c h\u1ec7 s\u1ed1 v\u00e0 $a\\ne 0$ <br\/>Do \u0111\u00f3: <br\/>1. Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+50x-1000=0$ l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u1edbi c\u00e1c h\u1ec7 s\u1ed1 $a=1;$ $b=50$ v\u00e0 $c=-1000$ <br\/>2. Ph\u01b0\u01a1ng tr\u00ecnh $1-2{{x}^{2}}=0$ l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u1edbi c\u00e1c h\u1ec7 s\u1ed1 $a=-2;$ $b=0$ v\u00e0 $c=1$ <br\/><br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng h\u00e0ng th\u1ee9 nh\u1ea5t l\u1ea7n l\u01b0\u1ee3t l\u00e0 $1;$ $50$ v\u00e0 $-1000$; \u00f4 tr\u1ed1ng h\u00e0ng th\u1ee9 hai l\u1ea7n l\u01b0\u1ee3t l\u00e0 $-2;$ $0$ v\u00e0 $1$<\/span><\/span>"}]}],"id_ques":821},{"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":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh n\u00e0o sau \u0111\u00e2y l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai?","select":["A. $4{{x}^{2}}-{{x}^{3}}+2=0$ ","B. $-2{{x}^{2}}=0$","C. $4x-5=0$ ","D. $2{{x}^{2}}-x\\left( 2x+1 \\right)=0$ "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai m\u1ed9t \u1ea9n l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 d\u1ea1ng $a{{x}^{2}}+bx+c=0$ trong \u0111\u00f3 $x$ l\u00e0 \u1ea9n; $a,b,c$ l\u00e0 c\u00e1c h\u1ec7 s\u1ed1 v\u00e0 $a\\ne 0$ <br\/>A. $4{{x}^{2}}-{{x}^{3}}+2=0$ c\u00f3 l\u0169y th\u1eeba b\u1eadc cao nh\u1ea5t l\u00e0 $3$ n\u00ean <b>kh\u00f4ng l\u00e0 <\/b> ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai.<br\/>B. $-2{{x}^{2}}=0$ <b>l\u00e0 <\/b> ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u1edbi h\u1ec7 s\u1ed1 $a=-2, b=c=0$<br\/>C. $4x-5=0$ <b>kh\u00f4ng l\u00e0 <\/b> ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u00ec h\u1ec7 s\u1ed1 $a=0$<br\/>D. $2{{x}^{2}}-x\\left( 2x+1 \\right)=0\\Leftrightarrow 2{{x}^{2}}-2{{x}^{2}}-x=0\\Leftrightarrow -x=0$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh $2{{x}^{2}}-x\\left( 2x+1 \\right)=0$ <b>kh\u00f4ng l\u00e0 <\/b> ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai v\u00ec h\u1ec7 s\u1ed1 $a=0$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":822},{"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":"<span class='basic_left'>\u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{3}{5}{{x}^{2}}+2x-7=3x+\\dfrac{1}{2}$ v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai . Khi \u0111\u00f3 h\u1ec7 s\u1ed1 $a,$ $b,$ $c$ c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0<\/span> ","select":["A. $a=\\dfrac{3}{5};b=-1;c=\\dfrac{15}{2}$ ","B. $a=\\dfrac{3}{5};b=-1;c=-\\dfrac{15}{2}$ ","C. $a=-\\dfrac{3}{5};b=-1;c=-\\dfrac{15}{2}$","D. $a=\\dfrac{3}{5};b=1;c=-\\dfrac{15}{2}$"],"hint":"Chuy\u1ec3n v\u1ebf v\u00e0 thu g\u1ecdn ph\u01b0\u01a1ng tr\u00ecnh \u0111\u01b0a v\u1ec1 d\u1ea1ng t\u1ed5ng qu\u00e1t $ax^2+bx+c=0$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & \\dfrac{3}{5}{{x}^{2}}+2x-7=3x+\\dfrac{1}{2} \\\\ & \\Leftrightarrow \\dfrac{3}{5}{{x}^{2}}+2x-7-3x-\\dfrac{1}{2}=0 \\\\ & \\Leftrightarrow \\dfrac{3}{5}{{x}^{2}}-x-\\dfrac{15}{2}=0 \\\\ \\end{align}$<br\/> V\u1eady $a=\\dfrac{3}{5};$ $b=-1;$ $c=-\\dfrac{15}{2}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":823},{"time":24,"part":[{"title":"\u0110i\u1ec1n 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 t\u1eadp nghi\u1ec7m l\u00e0 S={1;2;3} th\u00ec \u0111i\u1ec1n 1;2;3 ","temp":"fill_the_blank_random","correct":[[["0;2"],["2;0"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{2}}-2x=0$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S =$ {_input_} ","hint":"\u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng t\u00edch $a.b=0$ $\\Leftrightarrow a=0$ ho\u1eb7c $b=0$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned}& {{x}^{2}}-2x=0 \\\\ & \\Leftrightarrow x\\left( x-2 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x=2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S = \\left\\{0;2 \\right\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n \u0111i\u1ec1n l\u00e0 $0;2$ ho\u1eb7c $2;0$<\/span><\/span>"}]}],"id_ques":824},{"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\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $2{{x}^{2}}-1=0$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m l\u00e0 ","select":["A. $\\dfrac{\\sqrt{2}}{2}$ ","B. $\\dfrac{1}{2}$ v\u00e0 $-\\dfrac{1}{2}$","C. $\\dfrac{\\sqrt{2}}{2}$ v\u00e0 $-\\dfrac{\\sqrt{2}}{2}$ "],"hint":"\u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ${{A}^{2}}={{B}^{2}}\\Leftrightarrow \\left[ \\begin{aligned} & A=B \\\\ & A=-B \\\\ \\end{aligned} \\right.$ ","explain":"<span class='basic_left'>Chuy\u1ec3n v\u1ebf $-1$ v\u00e0 \u0111\u1ed5i d\u1ea5u c\u1ee7a n\u00f3, ta \u0111\u01b0\u1ee3c:<br\/> $\\begin{aligned} & 2{{x}^{2}}=1 \\\\ & \\Leftrightarrow {{x}^{2}}=\\dfrac{1}{2} \\\\ & \\Leftrightarrow x=\\pm \\dfrac{\\sqrt{2}}{2} \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=\\dfrac{\\sqrt{2}}{2}$ , ${{x}_{2}}=-\\dfrac{\\sqrt{2}}{2}$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":3}]}],"id_ques":825},{"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":[[["4"],["14"],["4"],["14"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-4x=-\\dfrac{1}{2}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=\\frac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}+\\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}}{2}$; ${{x}_{2}}=\\frac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}-\\sqrt{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}}{2}$ <br\/>v\u1edbi ${{x}_{1}}>{{x}_{2}}$","hint":"Th\u00eam v\u00e0o hai v\u1ebf c\u00f9ng m\u1ed9t s\u1ed1 \u0111\u1ec3 v\u1ebf tr\u00e1i th\u00e0nh b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t hi\u1ec7u.","explain":"<span class='basic_left'>T\u00e1ch $4x$ th\u00e0nh $2.x.2$ v\u00e0 th\u00eam $4$ v\u00e0o hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh, ta \u0111\u01b0\u1ee3c: <br\/>${{x}^{2}}-2.x.2+4=-\\dfrac{1}{2}+4$ <br\/>$\\begin{aligned} & \\Leftrightarrow {{\\left( x-2 \\right)}^{2}}=\\dfrac{7}{2} \\\\ & \\Leftrightarrow {{\\left( x-2 \\right)}^{2}}={{\\left( \\dfrac{\\sqrt{14}}{2} \\right)}^{2}} \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-2=\\dfrac{\\sqrt{14}}{2} \\\\ & x-2=-\\dfrac{\\sqrt{14}}{2} \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{\\sqrt{14}}{2}+2 \\\\ & x=-\\dfrac{\\sqrt{14}}{2}+2 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{4+\\sqrt{14}}{2} \\\\ & x=\\dfrac{4-\\sqrt{14}}{2} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m l\u00e0 ${{x}_{1}}=\\dfrac{4+\\sqrt{14}}{2}$ , ${{x}_{2}}=\\dfrac{4-\\sqrt{14}}{2}$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n theo th\u1ee9 t\u1ef1 t\u1eeb tr\u00e1i sang ph\u1ea3i l\u00e0:$4;14;4;14$ <\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span><br\/>C\u00e1ch gi\u1ea3i tr\u00ean nh\u1eb1m gi\u00fap h\u1ecdc sinh bi\u1ebft bi\u1ebfn \u0111\u1ed5i ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng t\u1ed5ng qu\u00e1t $ax^2+bx+c=0 (a \\ne 0)$ v\u1ec1 d\u1ea1ng<br\/> <\/span>${(x+\\dfrac{b}{2a})}^2=\\dfrac{b^2-4ac}{4a^2}$<br\/><span class='basic_left'>Sau khi c\u00e1c em h\u1ecdc c\u00f4ng th\u1ee9c nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai th\u00ec c\u00e1c em kh\u00f4ng c\u1ea7n bi\u1ebfn \u0111\u1ed5i nh\u01b0 tr\u00ean.<\/span>"}]}],"id_ques":826},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (<,>,=) th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["<"],[">"],["="]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ v\u1edbi $ a\\ne 0 $ v\u1edbi bi\u1ec7t th\u1ee9c $\\Delta$ <br\/>N\u1ebfu $\\Delta$ _input_0 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/>N\u1ebfu $\\Delta$ _input_0 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t<br\/>N\u1ebfu $\\Delta$ _input_0 th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p ","hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ v\u1edbi $ a\\ne 0 $ v\u1edbi bi\u1ec7t th\u1ee9c $\\Delta=b^2-4ac$ <br\/>N\u1ebfu $\\Delta<0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/>N\u1ebfu $\\Delta>0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t<br\/>N\u1ebfu $\\Delta=0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p <br\/><br\/><span class='basic_pink'>V\u1eady c\u00e1c d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $<;$ $>$ v\u00e0 $=.$<\/span><\/span>"}]}],"id_ques":827},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["121"],["25"],["1"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"<span class='basic_left'>Kh\u00f4ng gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh, t\u00ednh bi\u1ec7t th\u1ee9c $\\Delta $<br\/>1. Ph\u01b0\u01a1ng tr\u00ecnh $3{{x}^{2}}-5x-8=0$ c\u00f3 bi\u1ec7t th\u1ee9c $\\Delta =$_input_;<br\/>2. Ph\u01b0\u01a1ng tr\u00ecnh $-3{{x}^{2}}+7x-2=0$ c\u00f3 bi\u1ec7t th\u1ee9c $\\Delta =$_input_;<br\/>3. Ph\u01b0\u01a1ng tr\u00ecnh $6{{x}^{2}}+x=0$ c\u00f3 bi\u1ec7t th\u1ee9c $\\Delta =$_input_.","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/>B\u01b0\u1edbc 1: X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 $a,b,c$<br\/>B\u01b0\u1edbc 2: T\u00ednh bi\u1ec7t th\u1ee9c $\\Delta ={{b}^{2}}-4ac$ <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>1. Ph\u01b0\u01a1ng tr\u00ecnh $3{{x}^{2}}-5x-8=0$ c\u00f3 $a=3;b=-5;c=-8$<br\/>Ta c\u00f3 $\\Delta ={{b}^{2}}-4ac={{\\left( -5 \\right)}^{2}}-4.3.\\left( -8 \\right)=121$ <br\/>2. Ph\u01b0\u01a1ng tr\u00ecnh $-3{{x}^{2}}+7x-2=0$ c\u00f3 $a=-3;b=7;c=-2$<br\/>Ta c\u00f3 $\\Delta ={{b}^{2}}-4ac={{7}^{2}}-4.\\left( -3 \\right).\\left( -2 \\right)=25$<br\/>3. Ph\u01b0\u01a1ng tr\u00ecnh $6{{x}^{2}}+x=0$ c\u00f3 $a=6;b=1;c=0$<br\/>Ta c\u00f3 $\\Delta ={{b}^{2}}-4ac={{1}^{2}}-4.6.0=1$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $121;25;1.$<\/span><\/span>"}]}],"id_ques":828},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $x_{1} = 1; x_{2}= -\\dfrac{1}{3}$","B. $x_{1} = 1; x_{2}= \\dfrac{1}{3}$","C. $x_{1} = 1; x_{2}= -\\dfrac{1}{2}$"],"ques":"T\u00ecm nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh: $-3{{x}^{2}}+2x+1=0$<br\/>Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<br\/><br\/>$x_1=$?;$x_2=$?","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai b\u1eb1ng c\u00f4ng th\u1ee9c nghi\u1ec7m:<br\/>B\u01b0\u1edbc 1: X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 $a, b, c$ <br\/>B\u01b0\u1edbc 2: T\u00ednh $\\Delta $ <br\/>B\u01b0\u1edbc 3: \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai \u0111\u1ec3 t\u00ecm nghi\u1ec7m.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 l\u00e0 $a=-3;b=2;c=1$ <br\/>$\\Delta ={{2}^{2}}-4.\\left( -3 \\right).1=16.$ Suy ra $\\sqrt{\\Delta }=4$<br\/>Do $\\Delta >0,$ ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0:<br\/>${{x}_{1}}=\\dfrac{-2-4}{2.\\left( -3 \\right)}=1;$ ${{x}_{2}}=\\dfrac{-2+4}{2.\\left( -3 \\right)}=-\\dfrac{1}{3}$<br\/><\/span><span class='basic_green'>L\u01b0u \u00fd:<\/span>V\u1edbi nh\u1eefng ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai m\u00e0 h\u1ec7 s\u1ed1 $b$ c\u00f3 d\u1ea1ng $2m$ th\u00ec $b'=m$ th\u00ec ta n\u00ean d\u00f9ng c\u00f4ng th\u1ee9c nghi\u1ec7m thu g\u1ecdn.<\/span>"}]}],"id_ques":829},{"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":"<span class='basic_left'>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{2}}+x-1=0$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0","select":["A. $S=\\left\\{ \\dfrac{-1\\pm \\sqrt{3}}{2} \\right\\}$ ","B. $S=\\left\\{ \\dfrac{-1\\pm \\sqrt{5}}{2} \\right\\}$ ","C. $S=\\left\\{ \\dfrac{1\\pm \\sqrt{5}}{2} \\right\\}$ ","D. $S=\\varnothing$ "],"hint":"","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 l\u00e0 $a=1;b=1;c=-1$ <br\/>$\\Delta ={{1}^{2}}-4.1.\\left( -1 \\right)=5$ <br\/>Do $\\Delta >0$ , ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0:<br\/>${{x}_{1}}=\\dfrac{-1+\\sqrt{5}}{2};{{x}_{2}}=\\dfrac{-1-\\sqrt{5}}{2}$ <br\/> V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{ \\dfrac{-1\\pm \\sqrt{5}}{2} \\right\\}$<br\/> <span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":830},{"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"],["3"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"Ph\u01b0\u01a1ng tr\u00ecnh $9{{x}^{2}}-6x+1=0$ c\u00f3 nghi\u1ec7m l\u00e0 $x=\\frac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<br\/>(ph\u00e2n s\u1ed1 nh\u1eadn \u0111\u01b0\u1ee3c \u1edf d\u1ea1ng t\u1ed1i gi\u1ea3n)","hint":"","explain":"Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 l\u00e0 $a=9;b=-6;c=1$ <br\/>$\\Delta ={{\\left( -6 \\right)}^{2}}-4.9.1=0$ <br\/>Do $\\Delta =0$ , ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p<br\/>${{x}_{1}}={{x}_{2}}=\\dfrac{-\\left( -6 \\right)}{18}=\\dfrac{1}{3}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$ v\u00e0 $3.$<\/span>"}]}],"id_ques":831},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 nghi\u1ec7m th\u00edch h\u1ee3p v\u1edbi t\u1eebng ph\u01b0\u01a1ng tr\u00ecnh v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["0"],["2"]]],"list":[{"point":5,"width":40,"content":"","type_input":"","ques":"<span class='basic_left'>1. Ph\u01b0\u01a1ng tr\u00ecnh $16{{z}^{2}}+24z+9=0$ c\u00f3 _input_ nghi\u1ec7m<br\/>2. Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-x+3=0$ c\u00f3 _input_ nghi\u1ec7m<br\/>3. Ph\u01b0\u01a1ng tr\u00ecnh $-3{{x}^{2}}-5x+1=0$ c\u00f3 _input_ nghi\u1ec7m.","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i:<br\/>B\u01b0\u1edbc 1: X\u00e1c \u0111\u1ecbnh c\u00e1c h\u1ec7 s\u1ed1 $a, b, c$<br\/>B\u01b0\u1edbc 2: T\u00ednh bi\u1ec7t th\u1ee9c $\\Delta ={{b}^{2}}-4ac$ (ho\u1eb7c $\\Delta '=b{{'}^{2}}-ac$)<br\/>B\u01b0\u1edbc 3: + N\u1ebfu $\\Delta <0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/>+ N\u1ebfu $\\Delta =0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m<br\/>+ N\u1ebfu $\\Delta >0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>1. Ph\u01b0\u01a1ng tr\u00ecnh $16{{z}^{2}}+24z+9=0$ c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 l\u00e0 $a=16;b'=12;c=9$ <br\/>$\\Delta '={{12}^{2}}-16.9=0$ <br\/> Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 <b>1 <\/b>nghi\u1ec7m<br\/>2. Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-x+3=0$ c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 l\u00e0 $a=1;b=-1;c=3$ <br\/>$\\Delta ={{\\left( -1 \\right)}^{2}}-4.1.3=-11 <0$<br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh <b>v\u00f4 nghi\u1ec7m<\/b><br\/>3. Ph\u01b0\u01a1ng tr\u00ecnh $-3{{x}^{2}}-5x+1=0$ c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 l\u00e0 $a=-3;b=-5;c=1$ <br\/>$\\Delta ={{\\left( -5 \\right)}^{2}}-4.\\left( -3 \\right).1=37 >0$<br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 <b>2 <\/b>nghi\u1ec7m ph\u00e2n bi\u1ec7t.<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n theo th\u1ee9 t\u1ef1 l\u00e0 $1;0;2.$<\/span><\/span>"}]}],"id_ques":832},{"time":24,"part":[{"time":3,"title":"N\u1ed1i ph\u01b0\u01a1ng tr\u00ecnh \u1edf c\u1ed9t b\u00ean tr\u00e1i v\u1edbi bi\u1ec7t th\u1ee9c $\\Delta '$ t\u01b0\u01a1ng \u1ee9ng v\u1edbi n\u00f3 \u1edf c\u1ed9t b\u00ean ph\u1ea3i","title_trans":"","audio":"","temp":"matching","correct":[["3","1","4","2"]],"list":[{"point":5,"image":"img\/1.png","left":["$5{{x}^{2}}-6x+1=0$ ","$-2{{x}^{2}}+4\\sqrt{2}x-1=0$","${{x}^{2}}+\\dfrac{6}{5}x+2=0$ ","$4{{x}^{2}}+4x+1=0$ "],"right":["a. $\\Delta '=6$ ","b. $\\Delta '=0$","c. $\\Delta '=4$ ","d. $\\Delta '=-\\dfrac{41}{25}$ "],"top":60,"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>$b'=\\dfrac{b}{2}$ v\u00e0 $\\Delta '=b{{'}^{2}}-ac$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>1. Ph\u01b0\u01a1ng tr\u00ecnh $5{{x}^{2}}-6x+1=0$ c\u00f3 h\u1ec7 s\u1ed1 $a=5;b\u2019=-3;c=1$<br\/>$\\Delta '={{\\left( -3 \\right)}^{2}}-5.1=4$ <br\/>2. Ph\u01b0\u01a1ng tr\u00ecnh $-2{{x}^{2}}+4\\sqrt{2}x-1=0$ c\u00f3 h\u1ec7 s\u1ed1 $a=-2;$ $b'=2\\sqrt{2}$ ;$c=-1$<br\/>$\\Delta '={{\\left( 2\\sqrt{2} \\right)}^{2}}-\\left( -2 \\right).\\left( -1 \\right)=6$ <br\/>3. Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+\\dfrac{6}{5}x+2=0$ c\u00f3 h\u1ec7 s\u1ed1 $a=1;$ $b'=\\dfrac{3}{5}$ ;$c=2$<br\/>$\\Delta '={{\\left( \\dfrac{3}{5} \\right)}^{2}}-1.2=-\\dfrac{41}{25}$ <br\/>4. Ph\u01b0\u01a1ng tr\u00ecnh $4{{x}^{2}}+4x+1=0$ c\u00f3 h\u1ec7 s\u1ed1 $a=4; b\u2019=2;c=1$<br\/>$\\Delta '={{2}^{2}}-4.1=0$ <br\/><span class='basic_pink'>V\u1eady 1 n\u1ed1i v\u1edbi c, 2 n\u1ed1i v\u1edbi a, 3 n\u1ed1i v\u1edbi d v\u00e0 4 n\u1ed1i v\u1edbi b<\/span><\/span>"}]}],"id_ques":833},{"time":24,"part":[{"title":"Cho ph\u01b0\u01a1ng tr\u00ecnh $a{{x}^{2}}+bx+c=0$ v\u1edbi $a\\ne 0$ . ","title_trans":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","temp":"true_false","correct":[["f","f","f","t","f"]],"list":[{"point":5,"image":"img\/1.png","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["1. $a{{x}^{2}}+bx+c=0\\Leftrightarrow$${{\\left( x+\\dfrac{b}{2a} \\right)}^{2}}=\\dfrac{4ac-{{b}^{2}}}{4{{a}^{2}}}$ ","2. N\u1ebfu $\\Delta >0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ${{x}_{1}}=\\dfrac{-a+\\sqrt{\\Delta }}{2b};$ ${{x}_{2}}=\\dfrac{-a-\\sqrt{\\Delta }}{2b}$","3. N\u1ebfu $\\Delta >0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ${{x}_{1}}=\\dfrac{-c+\\sqrt{\\Delta }}{2a};$ ${{x}_{2}}=\\dfrac{-c-\\sqrt{\\Delta }}{2a}$","4. N\u1ebfu $\\Delta '>0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ${{x}_{1}}=\\dfrac{-b'+\\sqrt{\\Delta '}}{a};$ ${{x}_{2}}=\\dfrac{-b'-\\sqrt{\\Delta '}}{a}$","5. N\u1ebfu $\\Delta '=0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh nghi\u1ec7m k\u00e9p l\u00e0 ${{x}_{1}}={{x}_{2}}=-\\dfrac{b'}{2a}$ "],"hint":"","explain":["<span class='basic_left'>Sai v\u00ec<br\/> $a{{x}^{2}}+bx+c=0\\Leftrightarrow a\\left (x^2+2.x.\\dfrac{b}{2a}+\\dfrac{b^2}{4a^2}\\right)+c-\\dfrac{b^2}{4a}=0$<br\/>$\\Leftrightarrow a\\left (x+\\dfrac{b}{2a}\\right)^2=\\dfrac{b^2-4ac}{4a}$$\\Leftrightarrow{{\\left( x+\\dfrac{b}{2a} \\right)}^{2}}=\\dfrac{{{b}^{2}}-4ac}{4{{a}^{2}}}$<\/span>","<br\/><span class='basic_left'> Sai v\u00ec theo c\u00f4ng th\u1ee9c nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai:<br\/> N\u1ebfu $\\Delta >0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ${{x}_{1}}=\\dfrac{-b+\\sqrt{\\Delta }}{2a};$ ${{x}_{2}}=\\dfrac{-b-\\sqrt{\\Delta }}{2a}$<\/span>","<br\/><span class='basic_left'>Sai v\u00ec theo c\u00f4ng th\u1ee9c nghi\u1ec7m ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai:<br\/> N\u1ebfu $\\Delta >0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ${{x}_{1}}=\\dfrac{-b+\\sqrt{\\Delta }}{2a};$ ${{x}_{2}}=\\dfrac{-b-\\sqrt{\\Delta }}{2a}$<\/span>","<br\/><span class='basic_left'>\u0110\u00fang v\u00ec theo c\u00f4ng th\u1ee9c nghi\u1ec7m thu g\u1ecdn ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai:<br\/>N\u1ebfu $\\Delta '>0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ${{x}_{1}}=\\dfrac{-b'+\\sqrt{\\Delta '}}{a};$ ${{x}_{2}}=\\dfrac{-b'-\\sqrt{\\Delta '}}{a}$<\/span>","<br\/><span class='basic_left'> Sai v\u00ec theo c\u00f4ng th\u1ee9c nghi\u1ec7m thu g\u1ecdn ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai: N\u1ebfu $\\Delta '=0$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh nghi\u1ec7m k\u00e9p l\u00e0 ${{x}_{1}}={{x}_{2}}=-\\dfrac{b'}{a}$<\/span> "]}]}],"id_ques":834},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"],["-3"],["5"],["4"],["5"],["1"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: ${{x}^{2}}-6x+5=0$ b\u1eb1ng c\u00e1ch \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng: <br\/>$a=$_input_; $b'=$_input_; $c=$_input_<br\/><br\/>$\\Delta '=$_input_<br\/> Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:<br\/><br\/> $x_1=$_input_; $x_2=$_input_, bi\u1ebft $x_1>x_2$","hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn: <\/span><br\/>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai b\u1eb1ng c\u00f4ng th\u1ee9c nghi\u1ec7m:<br\/>B\u01b0\u1edbc 1: X\u00e1c \u0111\u1ecbnh h\u1ec7 s\u1ed1 $a, b', c$ <br\/>B\u01b0\u1edbc 2: T\u00ednh $\\Delta' $ <br\/>B\u01b0\u1edbc 3: \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c nghi\u1ec7m thu g\u1ecdn \u0111\u1ec3 t\u00ecm nghi\u1ec7m.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-6x+5=0$ c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 $a=1;b'=-3;c=5$<br\/>$\\Delta '={{\\left( -3 \\right)}^{2}}-5=4>0$<br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ${{x}_{1}}=3+2=5$ ; ${{x}_{2}}=3-2=1$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o c\u00e1c \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $1;-3;5;4;5;1.$<\/span><\/span>"}]}],"id_ques":835},{"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":"X\u00e1c \u0111\u1ecbnh $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-2\\left( m-1 \\right)x+{{m}^{2}}=0$ c\u00f3 nghi\u1ec7m k\u00e9p.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m=$?","hint":"Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p $\\Leftrightarrow \\Delta '=0$","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-2\\left( m-1 \\right)x+{{m}^{2}}=0$ c\u00f3 h\u1ec7 s\u1ed1 $a=1;b'=-\\left( m-1 \\right);c={{m}^{2}}$ <br\/>$\\begin{align} & \\Delta '={{\\left( m-1 \\right)}^{2}}-{{m}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,={{m}^{2}}-2m+1-{{m}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,=1-2m \\\\ \\end{align}$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p $\\Leftrightarrow \\Delta '=0\\Leftrightarrow 1-2m=0\\Leftrightarrow 2m=1\\Leftrightarrow m=\\dfrac{1}{2}$ <br\/>V\u1edbi $m=\\dfrac{1}{2},$ ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m k\u00e9p l\u00e0 ${{x}_{1}}={{x}_{2}}=m-1=-\\dfrac{1}{2}$ <br\/>V\u1eady $m=\\dfrac{1}{2}$ th\u00ec th\u1ecfa m\u00e3n y\u00eau c\u1ea7u \u0111\u1ec1 b\u00e0i.<\/span>"}]}],"id_ques":836},{"time":24,"part":[{"title":"\u0110i\u1ec1n d\u1ea5u (>,<,=) th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[[">"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"X\u00e1c \u0111\u1ecbnh $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-2\\left( m-1 \\right)x+{{m}^{2}}=0$ v\u00f4 nghi\u1ec7m.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m$_input_$\\dfrac{1}{2}$","hint":"Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m $\\Leftrightarrow \\Delta '<0$","explain":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-2\\left( m-1 \\right)x+{{m}^{2}}=0$ c\u00f3 h\u1ec7 s\u1ed1 $a=1;b'=-\\left( m-1 \\right);c={{m}^{2}}$ <br\/>$\\begin{align} & \\Delta '={{\\left( m-1 \\right)}^{2}}-{{m}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,={{m}^{2}}-2m+1-{{m}^{2}} \\\\ & \\,\\,\\,\\,\\,\\,=1-2m \\\\ \\end{align}$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m $\\Leftrightarrow \\Delta '<0\\Leftrightarrow 1-2m<0\\Leftrightarrow 2m>1\\Leftrightarrow m>\\dfrac{1}{2}$ <br\/>V\u1eady $m>\\dfrac{1}{2},$ ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m<br\/><span class='basic_pink'>V\u1eady d\u1ea5u c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $>$<\/span><\/span>"}]}],"id_ques":837},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"<span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh n\u00e0o sau \u0111\u00e2y c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t:<\/span>","select":["A. ${{x}^{2}}-5x+8=0$ ","B. $2{{x}^{2}}+1=0$","C. $3{{x}^{2}}-4x-7=0$ ","D. ${{x}^{2}}-x+\\dfrac{1}{4}=0$ "],"hint":"X\u00e1c \u0111\u1ecbnh s\u1ed1 nghi\u1ec7m c\u1ee7a t\u1eebng ph\u01b0\u01a1ng tr\u00ecnh","explain":"<span class='basic_left'>A. Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-5x+8=0$ c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 l\u00e0 $a=1;b=-5;c=8$ <br\/>$\\Delta ={{\\left( -5 \\right)}^{2}}-4.1.8=-7<0$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh <b>v\u00f4 nghi\u1ec7m<\/b><br\/>B. $2{{x}^{2}}+1=0\\Leftrightarrow 2{{x}^{2}}=-1$. Do v\u1ebf tr\u00e1i kh\u00f4ng \u00e2m, v\u1ebf ph\u1ea3i \u00e2m n\u00ean ph\u01b0\u01a1ng tr\u00ecnh <b>v\u00f4 nghi\u1ec7m<\/b><br\/>C. Ph\u01b0\u01a1ng tr\u00ecnh $3{{x}^{2}}-4x-7=0$ c\u00f3 c\u00e1c h\u1ec7 s\u1ed1 l\u00e0 $a=3;b=-4;c=-7$ <br\/>$\\Delta ={{\\left( -4 \\right)}^{2}}-4.3.\\left( -7 \\right)=100>0$ <br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 <b>2<\/b> nghi\u1ec7m ph\u00e2n bi\u1ec7t<br\/>D. ${{x}^{2}}-x+\\dfrac{1}{4}=0\\Leftrightarrow {{\\left( x-\\dfrac{1}{2} \\right)}^{2}}=0\\Leftrightarrow x-\\dfrac{1}{2}=0\\Leftrightarrow x=\\dfrac{1}{2}$<br\/>Suy ra ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 <b>1<\/b> nghi\u1ec7m<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":838},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o \u00f4 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-1"]]],"list":[{"point":5,"width":60,"type_input":"","ques":"T\u00ecm $a$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-ax-12=0$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=3$ .<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $a=$_input_","hint":"Thay $x=3$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec3 t\u00ecm $a.$","explain":"<span class='basic_left'>V\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=3$, n\u00ean ta thay $x=3$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\begin{aligned}& {{3}^{2}}-3a-12=0 \\\\ & \\Leftrightarrow -3a-3=0 \\\\ & \\Leftrightarrow a=-1 \\\\ \\end{aligned}$<br\/>V\u1edbi $a=-1,$ ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+x-12=0$ <br\/>$\\Delta ={{\\left( -1 \\right)}^{2}}-4.1.\\left( -12 \\right)=49 >0$<br\/>Khi \u0111\u00f3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 ${{x}_{1}}=\\dfrac{-1+7}{2}=3;{{x}_{2}}=\\dfrac{-1-7}{2}=-4$<br\/> V\u1eady $a=-1$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=3$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$<\/span><\/span>"}]}],"id_ques":839},{"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":"\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+2x-m=0$ c\u00f3 nghi\u1ec7m th\u00ec $m>-1$, <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?","select":["A. \u0110\u00fang ","B. Sai"],"hint":"","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc hai c\u00f3 nghi\u1ec7m $\\Leftrightarrow \\Delta ' \\ge 0$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i:<\/span><br\/>Ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}+2x-m=0$ c\u00f3 h\u1ec7 s\u1ed1 $a=1;b'=1;c=-m$ <br\/>$\\Delta '=1+m$ <br\/>Ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m $\\Leftrightarrow \\Delta ' \\ge 0\\Leftrightarrow m+1\\ge 0\\Leftrightarrow m\\ge -1$ <br\/>V\u1eady $m \\ge -1$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m.<br\/>Suy ra kh\u1eb3ng \u0111\u1ecbnh b\u00e0i to\u00e1n l\u00e0 sai.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n c\u1ea7n ch\u1ecdn l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":840}],"lesson":{"save":0,"level":1}}