{"segment":[{"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":[[["-1"],["3"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\left( 6-2y \\right)\\left( -1-y \\right)=0.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$","hint":"Ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 \u1edf d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch. \u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00edch \u0111\u1ec3 t\u00ecm nghi\u1ec7m.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & (6-2y)(-1-y)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 6-2y=0\\\\ & -1-y=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 2y=6 \\\\ &y=-1 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & y=3 \\\\ & y=-1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-1;3\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $-1$ v\u00e0 $3$<\/span><\/span>"}]}],"id_ques":861},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. {$\\dfrac{-2}{3};-2,2$}","B. {$\\dfrac{3}{2};1,2$}","C. {$\\dfrac{3}{2};-1,2$}"],"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh: $(2+3x)(-{{x}^{2}}+4)=0$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 S=?","hint":"\u00c1p d\u1ee5ng c\u00f4ng th\u1ee9c gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00edch \u0111\u1ec3 t\u00ecm nghi\u1ec7m.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & (2+3x)(-{{x}^{2}}+4)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 2+3x=0 \\\\ & -{{x}^{2}}+4=0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left[\\begin{aligned} & 3x=-2 \\\\ & {{x}^{2}}-4=0 \\\\ \\end{aligned} \\right. \\\\ &\\Leftrightarrow \\left[ \\begin{aligned} & x=\\frac{-2}{3} \\\\ & (x+2)(x-2)=0 \\\\\\end{aligned} \\right.\\Leftrightarrow \\left[ \\begin{aligned} & x=\\frac{-2}{3} \\\\ & x-2=0\\\\ & x+2=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{-2}{3} \\\\ & x=2 \\\\ & x=-2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{\\dfrac{-2}{3};-2;2\\right\\}$<\/span>"}]}],"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_random","correct":[[["1"],["-1"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $({{x}^{4}}-1)(2{{x}^{2}}+2)=0.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$","hint":"Ph\u00e2n t\u00edch \u0111a th\u1ee9c v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p s\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & ({{x}^{4}}-1)(2{{x}^{2}}+2)=0 \\\\ & \\Leftrightarrow \\left[\\begin{aligned} & {{x}^{4}}-1=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(1) \\\\ & 2{{x}^{2}}+2=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(2) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>Gi\u1ea3i (1). Ta c\u00f3:<br\/>$\\begin{aligned}& {{x}^{4}}-1=0 \\\\ & \\Leftrightarrow ({{x}^{2}}-1)({{x}^{2}}+1)=0 \\\\ & \\Leftrightarrow (x-1)(x+1)({{x}^{2}}+1)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-1=0 \\\\ & x+1=0 \\\\ & {{x}^{2}}+1=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=1 \\\\ & x=-1 \\\\ & {{x}^{2}}=-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{v\u00f4 nghi\u1ec7m v\u00ec } x^2 \\ge 0\\, \\forall\\,x) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 hai nghi\u1ec7m $x=1$ v\u00e0 $x=-1$(*)<br\/>Gi\u1ea3i (2). Ta c\u00f3:<br\/>$\\begin{aligned}& 2{{x}^{2}}+2=0 \\\\ & \\Leftrightarrow 2({{x}^{2}}+1)=0 \\\\ &\\Leftrightarrow {{x}^{2}}+1=0 \\\\ & \\Leftrightarrow {{x}^{2}}=-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{v\u00f4 nghi\u1ec7m v\u00ec } x^2 \\ge 0\\, \\forall\\,x) \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh (2) v\u00f4 nghi\u1ec7m. (**)<br\/>T\u1eeb (*) v\u00e0 (**), suy ra t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho l\u00e0 $S=\\{-1;1\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $-1$ v\u00e0 $1$<\/span><\/span>"}]}],"id_ques":863},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/8.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh $({{x}^{2}}-4)({{x}^{2}}+16)({{x}^{2}}+9)=0$ c\u00f3 $4$ nghi\u1ec7m.","select":["\u0110\u00fang","Sai"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00edch, t\u00ecm nghi\u1ec7m r\u1ed3i l\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n.","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & ({{x}^{2}}-4)({{x}^{2}}+16)({{x}^{2}}+9)=0 \\\\ &\\Leftrightarrow \\left[ \\begin{aligned}& {{x}^{2}}-4=0 \\\\ & {{x}^{2}}+16=0 \\\\ & {{x}^{2}}+9=0 \\\\ \\end{aligned} \\right.\\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & (x-2)(x+2)=0 \\\\ & {{x}^{2}}=-16\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{(v\u00f4 nghi\u1ec7m v\u00ec } x^2\\ge 0\\,\\forall x) \\\\ & {{x}^{2}}=-9\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{(v\u00f4 nghi\u1ec7m v\u00ec } x^2\\ge 0\\,\\forall x) \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=2 \\\\ & x=-2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m.<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai.<\/span><\/span>","column":2}]}],"id_ques":864},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/9.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh ${{\\left( 2x-5 \\right)}^{2}}({{x}^{2}}-9)({{x}^{4}}+1)=0$ c\u00f3 $4$ nghi\u1ec7m.","select":["\u0110\u00fang","Sai"],"hint":"","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & {{\\left( 2x-5 \\right)}^{2}}({{x}^{2}}-9)({{x}^{4}}+1)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{\\left( 2x-5 \\right)}^{2}}=0 \\\\ & {{x}^{2}}-9=0 \\\\ & {{x}^{4}}+1=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 2x-5=0 \\\\ & (x-3)(x+3)=0 \\\\ & {{x}^{4}}=-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{(v\u00f4 nghi\u1ec7m v\u00ec } x^4\\ge 0\\,\\forall x) \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 2x=5 \\\\ & x-3=0 \\\\ & x+3=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{5}{2} \\\\ & x=3 \\\\ & x=-3 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 ba nghi\u1ec7m.<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai.<\/span><\/span>","column":2}]}],"id_ques":865},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/10.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh $\\left( -{{x}^{2}}-3 \\right)\\left( {{x}^{2}}+5x+6 \\right)=0$ c\u00f3 $2$ nghi\u1ec7m.","select":["\u0110\u00fang","Sai"],"hint":"","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & \\left( -{{x}^{2}}-3 \\right)\\left( {{x}^{2}}+5x+6 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & -{{x}^{2}}-3=0 \\\\ & {{x}^{2}}+5x+6=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}+3=0 \\\\ & {{x}^{2}}+2x+3x+6=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & {{x}^{2}}=-3\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{v\u00f4 nghi\u1ec7m, v\u00ec}\\,\\, x^2\\ge 0\\,\\forall x) \\\\ & x(x+2)+3(x+2)=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow (x+3)(x+2)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x+3=0 \\\\ & x+2=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-3 \\\\ & x=-2 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 hai nghi\u1ec7m.<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}]}],"id_ques":866},{"time":24,"part":[{"title":"\u0110i\u1ec1n bi\u1ec3u th\u1ee9c th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2x-5","x+1","-5+2x"],["x+1","2x-5","-5+2x"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/11.jpg' \/><\/center><span class='basic_left'>\u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh $2{{x}^{2}}-3x-5=0$ v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> <br\/>$\\begin{align} & 2{{x}^{2}}-3x-5=0\\\\ & \\Leftrightarrow (\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})=0 \\\\ \\end{align}$<\/span>","hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch v\u00e0 nh\u00f3m \u0111\u1ec3 ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & 2{{x}^{2}}-3x-5=0 \\\\ & \\Leftrightarrow 2{{x}^{2}}-5x+2x-5=0 \\\\ & \\Leftrightarrow x(2x-5)+(2x-5)=0 \\\\ & \\Leftrightarrow (2x-5)(x+1)=0 \\\\ \\end{align}$<br\/><span class='basic_pink'>V\u1eady bi\u1ec3u th\u1ee9c c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $2x-5$ v\u00e0 $x+1$<\/span><\/span>"}]}],"id_ques":867},{"time":24,"part":[{"title":"\u0110i\u1ec1n bi\u1ec3u th\u1ee9c th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2x-5","2x-1","1-2x","5-2x"],["2x-1","2x-5","1-2x","5-2x"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/12.jpg' \/><\/center><span class='basic_left'>\u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh $4{{x}^{2}}-12x+5=0$ v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch.<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> <br\/>$\\begin{align} & 4{{x}^{2}}-12x+5=0\\\\ & \\Leftrightarrow (\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})(\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{})=0 \\\\ \\end{align}$<\/span>","hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p t\u00e1ch v\u00e0 nh\u00f3m \u0111\u1ec3 ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & 4{{x}^{2}}-12x+5=0 \\\\ & \\Leftrightarrow 4{{x}^{2}}-10x-2x+5=0\\\\ & \\Leftrightarrow 2x(2x-5)-(2x-5)=0 \\\\ & \\Leftrightarrow (2x-1)(2x-5)=0 \\\\ \\end{align}$<br\/><span class='basic_pink'>V\u1eady bi\u1ec3u th\u1ee9c c\u1ea7n \u0111i\u1ec1n v\u00e0o ch\u1ed7 tr\u1ed1ng l\u00e0 $2x-5$ v\u00e0 $2x-1$<\/span><\/span>"}]}],"id_ques":868},{"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,"width":50,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/13.jpg' \/><\/center><span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh ${{\\left( x-1 \\right)}^{3}}-{{(1-x)}^{2}}=0$ \u0111\u01b0a v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch c\u00f3 d\u1ea1ng:","select":["A. ${{(x-1)}^{2}}(x-3)=0$","B. ${{(x-1)}^{2}}(x-2)=0$","C. ${{(x-1)}^{2}}(2-x)=0$"],"hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t nh\u00e2n t\u1eed chung \u0111\u1ec3 ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & {{\\left( x-1 \\right)}^{3}}-{{(1-x)}^{2}}=0 \\\\ & \\Leftrightarrow{{(x-1)}^{3}}-{{(x-1)}^{2}}=0 \\\\ & \\Leftrightarrow {{(x-1)}^{2}}\\left[ (x-1)-1 \\right]=0 \\\\ & \\Leftrightarrow {{(x-1)}^{2}}(x-2)=0 \\\\ \\end{align}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":3}]}],"id_ques":869},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/1.png' \/><\/center><span class='basic_left'>B\u1ea1n H\u1ea3i gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-3x-4=2{{x}^{2}}-3x-5$ nh\u01b0 sau:<br\/>$\\begin{aligned} & {{x}^{2}}-3x-4=2{{x}^{2}}-3x-5 \\\\ & \\Leftrightarrow{{x}^{2}}+x-4x-4=2{{x}^{2}}+2x-5x-5 \\\\ & \\Leftrightarrow x(x+1)-4(x+1)=2x(x+1)-5(x+1) \\\\ & \\Leftrightarrow (x-4)(x+1)=(2x-5)(x+1) \\\\ & \\Leftrightarrow x-4=2x-5 \\\\ & \\Leftrightarrow x=1 \\\\ \\end{aligned}$<br\/> V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{1\\}$<br\/>B\u1ea1n H\u1ea3i gi\u1ea3i <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?<\/span>","select":["\u0110\u00fang","Sai"],"hint":"","explain":" <span class='basic_left'>B\u00e0i gi\u1ea3i c\u1ee7a H\u1ea3i l\u00e0 sai. <br\/>V\u00ec H\u1ea3i \u0111\u00e3 r\u00fat g\u1ecdn hai v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh cho $x+1$ khi ch\u01b0a x\u00e9t \u0111i\u1ec1u ki\u1ec7n $x+1\\ne 0$<br\/>B\u00e0i gi\u1ea3i \u0111\u00fang l\u00e0:<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & {{x}^{2}}-3x-4=2{{x}^{2}}-3x-5 \\\\ & \\Leftrightarrow {{x}^{2}}+x-4x-4=2{{x}^{2}}+2x-5x-5 \\\\ & \\Leftrightarrow x(x+1)-4(x+1)=2x(x+1)-5(x+1) \\\\ & \\Leftrightarrow (x-4)(x+1)=(2x-5)(x+1) \\\\ &\\Leftrightarrow (x-4)(x+1)-(2x-5)(x+1)=0 \\\\ & \\Leftrightarrow (x+1)\\left[ (x-4)-(2x-5) \\right]=0 \\\\ & \\Leftrightarrow (x+1)(-x+1)=0 \\\\ & \\Leftrightarrow \\left[\\begin{aligned} & x+1=0 \\\\ & -x+1=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-1 \\\\ & x=1 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-1;1\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span><br\/><span class='basic_green'>Sai l\u1ea7m th\u01b0\u1eddng g\u1eb7p: <\/span>Khi gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $A(x).B(x)=A(x)$, h\u1ecdc sinh th\u01b0\u1eddng r\u00fat g\u1ecdn hai v\u1ebf cho $A(x)$ khi ch\u01b0a c\u00f3 \u0111i\u1ec1u ki\u1ec7n $A(x) \\ne 0$. D\u1eabn \u0111\u1ebfn ph\u01b0\u01a1ng tr\u00ecnh gi\u1ea3i thi\u1ebfu nghi\u1ec7m.<br\/><i> <b>Ghi nh\u1edb:<\/b><\/i> V\u1edbi ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $A(x).B(x)=A(x)$, ta th\u1ef1c hi\u1ec7n chuy\u1ec3n v\u1ebf v\u00e0 \u0111\u1eb7t $A(x) $ l\u00e0m nh\u00e2n t\u1eed chung \u0111\u1ec3 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u00e2u v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch r\u1ed3i gi\u1ea3i ti\u1ebfp.<\/span>","column":2}]}],"id_ques":870},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/1.png' \/><\/center><span class='basic_left'>B\u1ea1n An gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{x}^{2}}-x-6=0$ nh\u01b0 sau:<br\/>$\\begin{aligned} & {{x}^{2}}-x=6 \\\\ & \\Leftrightarrow x(x-1)=2.3 \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & x=3 \\\\ & x-1=2 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left\\{ \\begin{aligned} & x=3 \\\\ & x=3 \\\\ \\end{aligned} \\right.\\\\ &\\Leftrightarrow x=3 \\\\ \\end{aligned}$<br\/> V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{3\\}$<br\/>B\u1ea1n An gi\u1ea3i <b>\u0111\u00fang<\/b> hay <b>sai<\/b>?<\/span>","select":["\u0110\u00fang","Sai"],"hint":"","explain":" <span class='basic_left'>B\u00e0i gi\u1ea3i c\u1ee7a An l\u00e0 sai. V\u00ec An \u0111\u00e3 s\u1eed d\u1ee5ng sai ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh trong t\u1eadp h\u1ee3p s\u1ed1 th\u1ef1c.<br\/>An c\u1ea7n chuy\u1ec3n v\u1ebf v\u00e0 ph\u00e2n t\u00edch \u0111a th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed \u0111\u1ec3 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch. <br\/>B\u00e0i gi\u1ea3i \u0111\u00fang l\u00e0:<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & {{x}^{2}}-x=6 \\\\ & \\Leftrightarrow {{x}^{2}}-x-6=0 \\\\ & \\Leftrightarrow {{x}^{2}}-3x+2x-6=0 \\\\ & \\Leftrightarrow x(x-3)+2(x-3)=0 \\\\ & \\Leftrightarrow (x+2)\\left( x-3 \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x+2=0 \\\\ & x-3=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-2 \\\\ & x=3 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-2;3\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 Sai.<\/span><br\/><span class='basic_green'>Nh\u1eadn x\u00e9t:<\/span> Trong t\u1eadp h\u1ee3p s\u1ed1 th\u1ef1c, khi gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $ax^2+bx=c$ , ta c\u1ea7n th\u1ef1c hi\u1ec7n chuy\u1ec3n h\u1eb1ng s\u1ed1 $c$ sang v\u1ebf tr\u00e1i, r\u1ed3i th\u1ef1c hi\u1ec7n ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed \u0111\u1ec3 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch r\u1ed3i gi\u1ea3i.<br\/><b>Tr\u00e1nh:<\/b> Gi\u1eef nguy\u00ean r\u1ed3i ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i \u0111\u1ed3ng nh\u1ea5t h\u1ec7 s\u1ed1. Ph\u01b0\u01a1ng ph\u00e1p n\u00e0y ch\u1ec9 s\u1eed d\u1ee5ng \u0111\u1ed1i v\u1edbi b\u00e0i to\u00e1n: T\u00ecm nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh trong t\u1eadp h\u1ee3p s\u1ed1 nguy\u00ean.<\/span>","column":2}]}],"id_ques":871},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/1.png' \/><\/center><span class='basic_left'>B\u1ea1n B\u00ecnh sau khi gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ${{(2x-3)}^{2}}={{(3-x)}^{2}}.$ B\u1ea1n th\u1ea5y thi\u1ebfu m\u1ea5t m\u1ed9t nghi\u1ec7m $x=0.$ H\u00e3y ch\u1ec9 ra cho b\u1ea1n B\u00ecnh bi\u1ebft, b\u1ea1n l\u00e0m <b>sai<\/b> t\u1eeb b\u01b0\u1edbc n\u00e0o? <br\/>B\u00e0i l\u00e0m c\u1ee7a b\u1ea1n B\u00ecnh:<br\/>$\\begin{align} & {{(2x-3)}^{2}}={{(3-x)}^{2}} \\\\ & \\Leftrightarrow 2x-3=3-x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,( \\text{ B\u01b0\u1edbc 1}) \\\\ & \\Leftrightarrow 2x-3-(3-x)=0\\,\\,\\,\\,\\,\\,( \\text{ B\u01b0\u1edbc 2}) \\\\ & \\Leftrightarrow 2x-3-3+x=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,( \\text{ B\u01b0\u1edbc 3}) \\\\ & \\Leftrightarrow 3x-6=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,( \\text{ B\u01b0\u1edbc 4}) \\\\ & \\Leftrightarrow x=2 \\\\ \\end{align}$<br\/> V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{2\\}$<\/span>","select":["A. B\u01b0\u1edbc 1","B. B\u01b0\u1edbc 2","C. B\u01b0\u1edbc 3","D. B\u01b0\u1edbc 4"],"hint":"","explain":" <span class='basic_left'>B\u00e0i l\u00e0m c\u1ee7a B\u00ecnh sai \u1edf b\u01b0\u1edbc 1, v\u00ec hai bi\u1ec3u th\u1ee9c $2x-3$ v\u00e0 $3-x$ ch\u01b0a d\u01b0\u01a1ng v\u1edbi m\u1ecdi $x$, n\u00ean khi khai c\u0103n hai v\u1ebf th\u00ec ph\u1ea3i \u0111\u1ec3 d\u1ea1ng tr\u1ecb tuy\u1ec7t \u0111\u1ed1i.<br\/>B\u00e0i gi\u1ea3i n\u00ean s\u1eeda l\u1ea1i nh\u01b0 sau:<br\/>$\\begin{aligned} & {{(2x-3)}^{2}}={{(3-x)}^{2}} \\\\ & \\Leftrightarrow {{(2x-3)}^{2}}-{{(3-x)}^{2}}=0 \\\\ & \\Leftrightarrow \\left[ 2x-3-(3-x) \\right]\\left[ 2x-3+(3-x) \\right]=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 2x-3-3+x=0 \\\\ & 2x-3+3-x=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 3x-6=0 \\\\ & x=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=2 \\\\ & x=0 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{0;2\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 A.<\/span><br\/><b>Nh\u1eadn x\u00e9t:<\/b> T\u1eeb b\u00e0i to\u00e1n tr\u00ean ta thu \u0111\u01b0\u1ee3c k\u1ebft qu\u1ea3: Khi gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $A^2(x)=B^2(x)$, ta lu\u00f4n c\u00f3 hai tr\u01b0\u1eddng h\u1ee3p $A(x)=B(x)$ ho\u1eb7c $A(x)=-B(x)$ khi ch\u01b0a bi\u1ebft d\u1ea5u c\u1ee7a $A(x)$ v\u00e0 $B(x)$.<\/span>","column":2}]}],"id_ques":872},{"time":24,"part":[{"title":"N\u1ed1i m\u1ed7i ph\u01b0\u01a1ng tr\u00ecnh c\u1ed9t tr\u00e1i v\u1edbi d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch t\u01b0\u01a1ng \u1ee9ng c\u1ee7a n\u00f3 \u1edf c\u1ed9t ph\u1ea3i. ","title_trans":"","audio":"","temp":"matching","correct":[["3","1","2","4"]],"list":[{"point":5,"image":"","left":["$A(x).B(x)=A(x)$ ","$A^2(x)=B^2(x)$","$A^3(x)=B^3(x)$ ","$A^4(x)-3A^2(x)+2=0$"],"right":["$[A(x)-B(x)][A(x)+B(x)]=0$","$[A(x)-B(x)][A^2(x)+A(x)B(x)+B^2(x)]=0$","$A(x)[B(x)-1]=0$","$[A^2(x)-1][A^2(x)-2]=0$"],"top":100,"hint":"","explain":"<span class='basic_left'>+)$A(x).B(x)=A(x) \\Leftrightarrow A(x).B(x)-A(x)=0 \\\\\\Leftrightarrow A(x)[B(x)-1]=0$<br\/>+) $A^2(x)=B^2(x)\\\\\\Leftrightarrow A^2(x)-B^2(x)=0 \\\\\\Leftrightarrow [A(x)-B(x)][A(x)+B(x)]=0$<br\/>+) $A^3(x)=B^3(x) \\\\\\Leftrightarrow A^3(x)-B^3(x)=0 \\\\\\Leftrightarrow [A(x)-B(x)][A^2(x)+A(x)B(x)+B^2(x)]=0$<br\/>+) $A^4(x)-3A^2(x)+2=0 \\\\\\Leftrightarrow A^4(x)-2A^2(x)-A^2(x)+2=0 \\\\\\Leftrightarrow A^2(x)[A^2(x)-2]-[A^2(x)-2]=0 \\\\\\Leftrightarrow [A^2(x)-1][A^2(x)-2]=0$<\/span>"}]}],"id_ques":873},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/13.jpg' \/><\/center>Cho hai bi\u1ec3u th\u1ee9c $A=9-6x+x^2$ v\u00e0 $B=4x^2+4x+1$. T\u00ecm $x$ \u0111\u1ec3 $A=B$.","select":["A. $x\\in\\left\\{\\dfrac {2}{3}\\right\\}$","B. $x\\in\\left\\{-4;\\dfrac {2}{3}\\right\\}$","C. $x\\in\\left\\{-2;1\\right\\}$","D. $x=-2$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A=B$. \u0110\u01b0a v\u1ebf tr\u00e1i v\u00e0 v\u1ebf ph\u1ea3i c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a m\u1ed9t t\u1ed5ng ho\u1eb7c m\u1ed9t hi\u1ec7u r\u1ed3i gi\u1ea3i ti\u1ebfp.","explain":" <span class='basic_left'><span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c bi\u1ebfn \u0111\u1ed5i m\u1ed7i v\u1ebf v\u1ec1 d\u1ea1ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $(a\\pm b)^2$.<br\/><b>B\u01b0\u1edbc 2:<\/b> Chuy\u1ec3n v\u1ebf v\u00e0 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng $A^2(x)-B^2(x)=0$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & A=B \\\\ & \\Leftrightarrow 9-6x+{{x}^{2}}=4{{x}^{2}}+4x+1 \\\\ & \\Leftrightarrow {{(3-x)}^{2}}={{(2x+1)}^{2}} \\\\ & \\Leftrightarrow {{(3-x)}^{2}}-{{(2x+1)}^{2}}=0 \\\\ & \\Leftrightarrow \\left[ (3-x)-(2x+1) \\right]\\left[ (3-x)+(2x+1) \\right]=0 \\\\ & \\Leftrightarrow \\left( 2-3x \\right)\\left( 4+x \\right)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 2-3x=0 \\\\ & 4+x=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{2}{3} \\\\ & x=-4 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\left\\{\\dfrac{2}{3};-4\\right\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><\/span>","column":2}]}],"id_ques":874},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/4.jpg' \/><\/center>Cho hai bi\u1ec3u th\u1ee9c $A= (x+1)^3 $ v\u00e0 $B =9x+9$. T\u00ecm $x$ \u0111\u1ec3 $A=B$.","select":["A. $x\\in\\left\\{-4;2\\right\\}$","B. $x\\in\\left\\{-1;2;4\\right\\}$","C. $x\\in\\left\\{-1;2;-4\\right\\}$","D. $x=2$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A=B$. ","explain":" <span class='basic_left'><span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1: <\/b>Chuy\u1ec3n v\u1ebf ph\u01b0\u01a1ng tr\u00ecnh $A=B$.<br\/><b>B\u01b0\u1edbc 2:<\/b> S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p \u0111\u1eb7t nh\u00e2n t\u1eed chung \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch.<br\/><b>B\u01b0\u1edbc 3:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh t\u00edch <br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><\/span> <br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & A=B \\\\ & \\Leftrightarrow {{(x+1)}^{3}}=9x+9 \\\\ & \\Leftrightarrow {{(x+1)}^{3}}=9(x+1) \\\\ & \\Leftrightarrow {{(x+1)}^{3}}-9(x+1)=0 \\\\ & \\Leftrightarrow (x+1)\\left[ {{(x+1)}^{2}}-9 \\right]=0 \\\\ & \\Leftrightarrow (x+1)\\left[ (x+1)-3 \\right]\\left[ (x+1)+3 \\right]=0 \\\\ & \\Leftrightarrow (x+1)(x-2)(x+4)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x+1=0 \\\\ & x-2=0 \\\\ & x+4=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-1 \\\\ & x=2 \\\\ & x=-4 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-1;2;-4\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span>","column":2}]}],"id_ques":875},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/2.jpg' \/><\/center>Cho hai bi\u1ec3u th\u1ee9c $A=x^3$ v\u00e0 $B =6x^2$. T\u00ecm $x$ \u0111\u1ec3 $A-B=-9x$.","select":["A. $x=3$","B. $x\\in\\left\\{-3;0\\right\\}$","C. $x=0$","D. $x\\in\\left\\{0;3\\right\\}$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A-B=-9x$. Chuy\u1ec3n v\u1ebf r\u1ed3i ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed.","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned}& A-B=-9x \\\\ & \\Leftrightarrow {{x}^{3}}-6{{x}^{2}}=-9x \\\\ & \\Leftrightarrow {{x}^{3}}-6{{x}^{2}}+9x=0 \\\\ & \\Leftrightarrow x({{x}^{2}}-6x+9)=0 \\\\ & \\Leftrightarrow x{{(x-3)}^{2}}=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & {{(x-3)}^{2}}=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x-3=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x=3 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{0;3\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span>","column":2}]}],"id_ques":876},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/10.jpg' \/><\/center>Cho hai bi\u1ec3u th\u1ee9c $A= x-3 $ v\u00e0 $B=x+3$. T\u00ecm $x$ \u0111\u1ec3 $A.B=16$.","select":["A. $x\\in\\left\\{-5;5\\right\\}$","B. $x=5$","C. $x=-3$","D. $x\\in\\left\\{-3;3\\right\\}$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A.B=16$. \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c bi\u1ebfn \u0111\u1ed5i v\u1ebf tr\u00e1i, chuy\u1ec3n v\u1ebf v\u00e0 t\u00ecm $x$","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{aligned} & A.B=16 \\\\ & \\Leftrightarrow (x-3)(x+3)=16 \\\\ & \\Leftrightarrow {{x}^{2}}-9=16 \\\\ & \\Leftrightarrow {{x}^{2}}-25=0 \\\\ & \\Leftrightarrow (x-5)(x+5)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x-5=0 \\\\ & x+5=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-5 \\\\ & x=5 \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-5;5\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":877},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/12.jpg' \/><\/center>Cho hai bi\u1ec3u th\u1ee9c $A=x^4+1$ v\u00e0 $B=2x^2$. T\u00ecm $x$ \u0111\u1ec3 $A=B$.","select":["A. $x=1$","B. $x\\in\\left\\{-1;1\\right\\}$","C. $x=-1$","D. $x\\in \\varnothing$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A=B$ b\u1eb1ng c\u00e1ch: \u0110\u1eb7t $x^2=t$. ","explain":" <span class='basic_left'><span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Chuy\u1ec3n v\u1ebf, \u0111\u1eb7t $x^2=t$<br\/><b>B\u01b0\u1edbc 2:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$<br\/><b>B\u01b0\u1edbc 3:<\/b> Thay gi\u00e1 tr\u1ecb c\u1ee7a $t$ t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o $x^2=t$ \u0111\u1ec3 t\u00ecm $x$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><\/span>Ta c\u00f3:<br\/>$\\begin{align} & A=B \\\\ & \\Leftrightarrow {{x}^{4}}+1=2{{x}^{2}} \\\\ & \\Leftrightarrow {{x}^{4}}-2{{x}^{2}}+1=0 \\\\ \\end{align}$<br\/>\u0110\u1eb7t $x^2=t\\,\\,(t\\ge 0)$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh:<br\/>$\\begin{align} & {{t}^{2}}-2t+1=0 \\\\ & \\Leftrightarrow {{(t-1)}^{2}}=0 \\\\ & \\Leftrightarrow t-1=0 \\\\ & \\Leftrightarrow t=1(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{align}$<br\/>V\u1edbi $t=1$, suy ra $x^2=1 \\Leftrightarrow x=1$ ho\u1eb7c $x=-1$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-1;1\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B.<\/span><br\/><span class='basic_green'>Ghi nh\u1edb: <\/span> Ph\u01b0\u01a1ng tr\u00ecnh c\u1ee7a b\u00e0i to\u00e1n tr\u00ean \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng, c\u00f3 d\u1ea1ng t\u1ed5ng qu\u00e1t l\u00e0: $ax^4+bx^2+c=0$.(1)<br\/>Ph\u01b0\u01a1ng ph\u00e1p gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh tr\u00f9ng ph\u01b0\u01a1ng:<br\/>+) \u0110\u1eb7t $x^2=t\\,\\,\\,(t\\ge 0)$. (*)<br\/>+) Ph\u01b0\u01a1ng tr\u00ecnh (1) tr\u1edf th\u00e0nh: $at^2+bt+c=0$ (2) l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc 2 \u1ea9n $t$. <br\/>+) \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh (2) v\u1ec1 ph\u01b0\u01a1ng tr\u00ecnh t\u00edch \u0111\u1ec3 t\u00ecm $t$.<br\/>+) Thay gi\u00e1 tr\u1ecb $t$ t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o (*) \u0111\u1ec3 t\u00ecm $x$.<\/span>","column":2}]}],"id_ques":878},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/2.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x^4+3x^2+2=0$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: ","select":["A. $S=\\{1\\}$","B. $S=\\left\\{-1;1\\right\\}$","C. $S=\\{-1\\}$","D. $S= \\varnothing$"],"hint":"\u0110\u1eb7t $x^2=t$. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$ r\u1ed3i t\u00ecm $x$","explain":" <span class='basic_left'><span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> \u0110\u1eb7t $x^2=t$<br\/><b>B\u01b0\u1edbc 2:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$<br\/><b>B\u01b0\u1edbc 3:<\/b> Thay gi\u00e1 tr\u1ecb c\u1ee7a $t$ t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o $x^2=t$ \u0111\u1ec3 t\u00ecm $x$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><\/span>Ta c\u00f3:<br\/>\u0110\u1eb7t $x^2=t\\,\\,(t\\ge 0)$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh:<br\/>$\\begin{aligned} & {{t}^{2}}+3t+2=0 \\,\\,\\,(1)\\\\ & \\Leftrightarrow {{t}^{2}}+2t+t+2=0 \\\\ & \\Leftrightarrow t(t+2)+(t+2)=0 \\\\ & \\Leftrightarrow (t+1)(t+2)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t+1=0 \\\\ & t+2=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t=-1\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i )} \\\\ & t=-2\\,\\,\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i)} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh (1) v\u00f4 nghi\u1ec7m.<br\/> Suy ra, ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 t\u1eadp nghi\u1ec7m $S=\\varnothing$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D.<\/span><\/span>","column":2}]}],"id_ques":879},{"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":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai19/lv2/img\/2.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x^4-5x^2+6=0$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0: ","select":["A. $S=\\{-\\sqrt {2};\\sqrt {2};-\\sqrt {3};\\sqrt {3}\\}$","B. $S=\\{\\sqrt {2};\\sqrt {3}\\}$","C. $S=\\{{2};{3}\\}$","D. $S= \\varnothing$"],"hint":"\u0110\u1eb7t $x^2=t$. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$ r\u1ed3i t\u00ecm $x$","explain":" <span class='basic_left'><span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> \u0110\u1eb7t $x^2=t$<br\/><b>B\u01b0\u1edbc 2:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$<br\/><b>B\u01b0\u1edbc 3:<\/b> Thay gi\u00e1 tr\u1ecb c\u1ee7a $t$ t\u00ecm \u0111\u01b0\u1ee3c v\u00e0o $x^2=t$ \u0111\u1ec3 t\u00ecm $x$<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><\/span>Ta c\u00f3:<br\/>\u0110\u1eb7t $x^2=t\\,\\,(t\\ge 0)$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh:<br\/>$\\begin{aligned} & {{t}^{2}}-5t+6=0 \\,\\,\\,(1)\\\\ & \\Leftrightarrow {{t}^{2}}-2t-3t+6=0 \\\\ & \\Leftrightarrow t(t-2)-3(t-2)=0 \\\\ & \\Leftrightarrow (t-2)(t-3)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t-2=0 \\\\ & t-3=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & t=2\\\\ & t=3\\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1edbi $t=2$, ta c\u00f3:<br\/>$x^2=2 \\Leftrightarrow x^2-2=0 \\Leftrightarrow (x-\\sqrt {2})(x+\\sqrt {2})=0 \\Leftrightarrow \\left[ \\begin{aligned} & x=\\sqrt {2} \\\\ & x=-\\sqrt {2} \\\\ \\end{aligned} \\right.$<br\/>V\u1edbi $t=3$, ta c\u00f3:<br\/>$x^2=3 \\Leftrightarrow x^2-3=0 \\Leftrightarrow (x-\\sqrt {3})(x+\\sqrt {3})=0 \\Leftrightarrow \\left[ \\begin{aligned} & x=\\sqrt {3} \\\\ & x=-\\sqrt {3} \\\\ \\end{aligned} \\right.$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-\\sqrt {2};\\sqrt {2};-\\sqrt {3};\\sqrt {3}\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A.<\/span><\/span>","column":2}]}],"id_ques":880}],"lesson":{"save":0,"level":2}}