{"segment":[{"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $(x+1)^2-2(x+1)(3-2x)+(2x-3)^2$. Ta \u0111\u01b0\u1ee3c:","select":["A. $(x+4)^2$","B. $(x-4)^2$","C. $(2x-3)^2$","D. $(3x-2)^2$"],"hint":"","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$(x+1)^2-2(x+1)(3-2x)+(2x-3)^2\\\\ = (x+1)^2-2(x+1)(3-2x)+(3-2x)^2\\\\ = [(x+1)-(3-2x)]^2\\\\ = (x+1-3+2x)^2\\\\ =(3x-2)^2$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}],"id_ques":261},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Ph\u00e2n t\u00edch \u0111a th\u1ee9c $4x^2-25+(x+7)(5-2x)$ th\u00e0nh nh\u00e2n t\u1eed. Ta \u0111\u01b0\u1ee3c:","select":["A. $(2x-5)(x+12)$","B. $(2x-5)(3x+12)$","C. $(2x-5)(x-2)$","D. $(2x+5)(x-2)$"],"hint":"","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$4x^2-25+(x+7)(5-2x)\\\\ =(2x)^2-5^2+(x+7)(5-2x)\\\\ =(2x-5)(2x+5)-(x+7)(2x-5)\\\\ =(2x-5)[(2x+5)-(x+7)]\\\\ =(2x-5)(x-2)$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}],"id_ques":262},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank_random","correct":[[["0"],["1"],["2"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"T\u00ecm $x$ bi\u1ebft $x^3-3x^2+2x=0$.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $x\\in \\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$","hint":"","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$x^3-3x^2+2x=0\\\\ \\Leftrightarrow x(x^2-3x+2)=0\\\\ \\Leftrightarrow x(x^2-2x-x+2)=0\\\\ \\Leftrightarrow x[x(x-2)-(x-2)]=0\\\\ \\Leftrightarrow x(x-2)(x-1)=0 \\\\ \\Leftrightarrow \\left[\\begin{align}&x=0\\\\ &x=1 \\\\ &x=2 \\\\ \\end{align}\\right.$<br\/>V\u1eady $x\\in \\{0;1;2\\}$<br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0;\\,1;\\,2$<\/span><\/span>"}],"id_ques":263},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 d\u01b0\u1edbi d\u1ea1ng ph\u00e2n s\u1ed1 t\u1ed1i gi\u1ea3n","temp":"fill_the_blank","correct":[[["-5"],["3"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"frac","ques":"T\u00ecm $x$ bi\u1ebft $(3x+2)(3x-2)-(3x+1)^2=5$.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $x=$<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$(3x+2)(3x-2)-(3x+1)^2=5\\\\ \\Leftrightarrow (3x)^2-4-(3x+1)^2-5=0\\\\ \\Leftrightarrow 9x^2-4-(9x^2+6x+1)-5=0\\\\ \\Leftrightarrow 9x^2-4-9x^2-6x-1-5=0\\\\ \\Leftrightarrow -6x-10=0\\\\ \\Leftrightarrow x=\\dfrac{-5}{3} $<\/span>"}],"id_ques":264},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Gi\u00e1 tr\u1ecb bi\u1ec3u th\u1ee9c $P=(x+1)^3-(x-1)^3-3[(x-1)^2+(x+1)^2]$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o gi\u00e1 tr\u1ecb c\u1ee7a $x$. <b>\u0110\u00fang<\/b> hay <b>Sai<\/b>?","select":[" \u0110\u00fang "," Sai "],"hint":"","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$P=(x+1)^3-(x-1)^3-3[(x-1)^2+(x+1)^2]\\\\ \\,\\,\\,\\,\\,=x^3+3x^2+3x+1-(x^3-3x^2+3x-1)-3(x^2-2x+1+x^2+2x+1)\\\\ \\,\\,\\,\\,\\,=x^3+3x^2+3x+1-x^3+3x^2-3x+1-3(2x^2+2)\\\\ \\,\\,\\,\\,\\,=6x^2+2-6x^2-6\\\\ \\,\\,\\,\\,\\,=-4$<br\/>Suy ra, $P=-4$ v\u1edbi m\u1ecdi $x$<br\/>Do \u0111\u00f3, gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c $P$ kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o gi\u00e1 tr\u1ecb c\u1ee7a $x$.<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}],"id_ques":265},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110i\u1ec1n k\u1ebft qu\u1ea3 d\u01b0\u1edbi d\u1ea1ng ph\u00e2n s\u1ed1 t\u1ed1i gi\u1ea3n","temp":"fill_the_blank","correct":[[["27"],["4"],["3"],["2"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"frac","ques":"T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $A= x^2-3x+9$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $Min\\,A=$<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>khi $x=$<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $A$ v\u1ec1 d\u1ea1ng: $f^2(x)+m$ v\u1edbi $m$ l\u00e0 h\u1eb1ng s\u1ed1, r\u1ed3i \u0111\u00e1nh gi\u00e1.","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$A=x^2-3x+9=x^2-2.x.\\dfrac{3}{2}+\\dfrac{9}{4}+\\dfrac{27}{4}=\\left(x-\\dfrac{3}{2}\\right)^2+\\dfrac{27}{4}$<br\/>V\u00ec $\\left(x-\\dfrac{3}{2}\\right)^2 \\ge\\,0 \\,\\,\\forall x\\in\\mathbb R$ n\u00ean $A=\\left(x-\\dfrac{3}{2}\\right)^2+\\dfrac{27}{4}\\ge\\,\\dfrac{27}{4}\\,\\forall x $<br\/>V\u1eady gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $A$ l\u00e0 $\\dfrac{27}{4}$.<br\/>D\u1ea5u \"=\" x\u1ea3y ra khi $\\left(x-\\dfrac{3}{2}\\right)^2=0\\Leftrightarrow x=\\dfrac{3}{2}$<\/span>"}],"id_ques":266},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"Cho hai bi\u1ec3u th\u1ee9c $A=x^4-x^3+6x^2-x+a$ v\u00e0 $B=x^2-x+5$. T\u00ecm $a$ \u0111\u1ec3 $A$ chia cho $B$ \u0111\u01b0\u1ee3c s\u1ed1 d\u01b0 l\u00e0 \u0111a th\u1ee9c b\u1eadc nh\u1ea5t h\u1ec7 s\u1ed1 t\u1ef1 do l\u00e0 $2$.","select":["A. $a=-22$","B. $a=22$","C. $a=-20$","D. $a=20$"],"hint":"Th\u1ef1c hi\u1ec7n ph\u00e9p chia v\u00e0 t\u00ecm s\u1ed1 d\u01b0.","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$\\left. \\begin{aligned} & \\begin{matrix} \\,\\,\\,\\,\\,{{x}^{4}}-2{{x}^{3}}+10{{x}^{2}}-x+a\\,\\,\\,\\,\\,\\,\\,\\, \\\\ \\,\\,\\,\\,\\,\\,{{x}^{4}}-{{x}^{3}}\\,\\,+5{{x}^{2}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\\\\\end{matrix} \\\\ & \\overline{\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-{{x}^{3}}+5{{x}^{2}}-x+a\\,\\, \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-{{x}^{3}}+{{x}^{2}}-5x \\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\begin{aligned} & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4{{x}^{2}}+4x+a\\\\ & \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,4{{x}^{2}}-4x+20\\,\\, \\\\ &\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\overline{\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,8x+a-20}\\, \\\\ \\end{aligned}} \\\\ \\end{aligned}} \\\\ \\end{aligned} \\right|\\begin{matrix} \\dfrac{{{x}^{2}}-x+5}{{{x}^{2}}-x+4} \\\\ \\begin{matrix} \\begin{matrix} \\begin{matrix} {} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix} \\\\ {} \\\\\\end{matrix}$<br\/>V\u1eady d\u01b0 c\u1ee7a ph\u00e9p chia $A$ cho $B$ l\u00e0 $8x+a-20$<br\/>\u0110\u1ec3 s\u1ed1 d\u01b0 l\u00e0 \u0111a th\u1ee9c b\u1eadc nh\u1ea5t h\u1ec7 s\u1ed1 t\u1ef1 do b\u1eb1ng $2$ th\u00ec $a-20=2\\Leftrightarrow a=22$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}],"id_ques":267},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[4]],"list":[{"point":5,"ques":"R\u00fat g\u1ecdn bi\u1ec3u th\u1ee9c $A=\\dfrac{2x}{x-3}-\\dfrac{x+1}{x+3}+\\dfrac{{{x}^{2}}-1}{9-{{x}^{2}}}$","select":["A. $\\dfrac{6x+4}{(x-3)(x+3)}$","B. $\\dfrac{2x^2+8x+4}{(x-3)(x+3)}$","C. $\\dfrac{4}{x+3}$","D. $\\dfrac{8x+4}{(x-3)(x+3)}$"],"hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 3$ v\u00e0 $x\\ne -3$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & A=\\dfrac{2x}{x-3}-\\dfrac{x+1}{x+3}+\\dfrac{{{x}^{2}}-1}{9-{{x}^{2}}} \\\\ & \\,\\,\\,\\,=\\dfrac{2x}{x-3}-\\dfrac{x+1}{x+3}-\\dfrac{{{x}^{2}}-1}{{{x}^{2}}-9} \\\\ & \\,\\,\\,\\,=\\dfrac{2x(x+3)}{(x-3)(x+3)}-\\dfrac{(x+1)(x-3)}{(x-3)(x+3)}-\\dfrac{{{x}^{2}}-1}{(x-3)(x+3)} \\\\ & \\,\\,\\,\\,=\\dfrac{2{{x}^{2}}+6x-({{x}^{2}}-2x-3)-({{x}^{2}}-1)}{(x-3)(x+3)} \\\\ & \\,\\,\\,\\,=\\dfrac{2{{x}^{2}}+6x-{{x}^{2}}+2x+3-{{x}^{2}}+1}{(x-3)(x+3)} \\\\ & \\,\\,\\,\\,=\\dfrac{8x+4}{(x-3)(x+3)} \\\\ \\end{aligned}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}],"id_ques":268},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank","correct":[[["7"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $B=\\left( \\dfrac{2x}{x-2}-\\dfrac{x+1}{x+2}+\\dfrac{{{x}^{2}}-x-1}{4-{{x}^{2}}} \\right):\\left( 1-\\dfrac{x-1}{x+2} \\right)$<br\/><b> C\u00e2u a<\/b>: T\u00ecm $x$ khi $B=3$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $x=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span>","hint":"R\u00fat g\u1ecdn $B$ r\u1ed3i gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $B=3.$","explain":"<span class='basic_left'><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh $x\\ne \\pm 2$<br\/>Ta c\u00f3: <br\/>$\\begin{aligned} & B=\\left( \\dfrac{2x}{x-2}-\\dfrac{x+1}{x+2}+\\dfrac{{{x}^{2}}-x-1}{4-{{x}^{2}}} \\right):\\left( 1-\\dfrac{x-1}{x+2} \\right) \\\\ & \\,\\,\\,\\,=\\left[ \\dfrac{2x(x+2)}{(x-2)(x+2)}-\\dfrac{(x+1)(x-2)}{(x+2)(x-2)}-\\dfrac{{{x}^{2}}-x-1}{{{x}^{2}}-4} \\right]:\\left[ \\dfrac{(x+2)-(x-1)}{x+2} \\right] \\\\ & \\,\\,\\,\\,=\\left[ \\dfrac{2{{x}^{2}}+4x}{(x-2)(x+2)}-\\dfrac{{{x}^{2}}-x-2}{(x-2)(x+2)}-\\dfrac{{{x}^{2}}-x-1}{{{x}^{2}}-4} \\right]:\\left( \\dfrac{x+2-x+1}{x+2} \\right) \\\\ & \\,\\,\\,\\,=\\dfrac{2{{x}^{2}}+4x-{{x}^{2}}+x+2-{{x}^{2}}+x+1}{(x-2)(x+2)}:\\dfrac{3}{x+2} \\\\ & \\,\\,\\,\\,=\\dfrac{6x+3}{(x-2)(x+2)}.\\dfrac{x+2}{3} \\\\ & \\,\\,\\,\\,=\\dfrac{6x+3}{3(x-2)}\\\\ & \\,\\,\\,\\,=\\dfrac{2x+1}{x-2} \\\\ \\end{aligned}$<br\/>V\u1edbi $B=3$, ta c\u00f3:<br\/>$\\dfrac{2x+1}{x-2}=3 \\Rightarrow 2x+1=3(x-2)\\Leftrightarrow 2x+1=3x-6\\Leftrightarrow x=7$ (th\u1ecfa m\u00e3n)<br\/>V\u1eady $x=7$ th\u00ec $B=3$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $7$<\/span><\/span>"}],"id_ques":269},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"fill_the_blank_random","correct":[[["-3"],["1"],["3"],["7"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $B=\\left( \\dfrac{2x}{x-2}-\\dfrac{x+1}{x+2}+\\dfrac{{{x}^{2}}-x-1}{4-{{x}^{2}}} \\right):\\left( 1-\\dfrac{x-1}{x+2} \\right)$<br\/><b> C\u00e2u b<\/b>: T\u00ecm $x \\in \\mathbb Z$ \u0111\u1ec3 $B\\in \\mathbb Z$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $x\\in\\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$<\/span>","hint":"","explain":"<span class='basic_left'><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh $x\\ne \\pm 2$<br\/>Theo c\u00e2u a, ta c\u00f3: $B=\\dfrac{2x+1}{x-2}\\\\ \\Leftrightarrow B=\\dfrac{2(x-2)+5}{x-2}\\\\ \\Leftrightarrow B=2+\\dfrac{5}{x-2}$<br\/>V\u1eady \u0111\u1ec3 $B\\in \\mathbb Z$ th\u00ec $x-2 \\in \u01af(5)=\\{\\pm 1; \\pm 5\\}$<br\/>Ta c\u00f3 b\u1ea3ng:<br\/><table><tr><td>$x-2$<\/td><td>$-5$<\/td><td>$-1$<\/td><td>$1$<\/td><td>$5$<\/td><\/tr><tr><td>$x$<\/td><td>$-3$<\/td><td>$1$<\/td><td>$3$<\/td><td>$7$<\/td><\/tr><\/table><br\/>$x\\in \\{-3;1;3;7\\}$ th\u00ec $B\\in \\mathbb Z$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-3;1;3;7$<\/span><\/span>"}],"id_ques":270},{"title_trans":"","title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. $\\dfrac{3}{x-2}$","B. $\\dfrac{1}{x+3}$","C. $\\dfrac{2}{x+2}$"],"ques":"Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{{{x}^{2}}-3x}{{{x}^{2}}-9}-1 \\right):\\left( \\dfrac{9-{{x}^{2}}}{{{x}^{2}}+x-6}-\\dfrac{x-3}{2-x}-\\dfrac{x-2}{x+3} \\right)$<br\/><b> C\u00e2u a<\/b>: R\u00fat g\u1ecdn $C$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $C=$?","hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne \\{-3;2;3\\}$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & C=\\left( \\dfrac{{{x}^{2}}-3x}{{{x}^{2}}-9}-1 \\right):\\left( \\dfrac{9-{{x}^{2}}}{{{x}^{2}}+x-6}-\\dfrac{x-3}{2-x}-\\dfrac{x-2}{x+3} \\right) \\\\ & \\,\\,\\,\\,=\\left[ \\dfrac{{{x}^{2}}-3x}{(x-3)(x+3)}-1 \\right]:\\left[ \\dfrac{9-{{x}^{2}}}{(x-2)(x+3)}+\\dfrac{x-3}{x-2}-\\dfrac{x-2}{x+3} \\right] \\\\ & \\,\\,\\,\\,=\\left[ \\dfrac{{{x}^{2}}-3x}{(x-3)(x+3)}-\\dfrac{{{x}^{2}}-9}{(x-3)(x+3)} \\right]:\\left[ \\dfrac{9-{{x}^{2}}}{(x-2)(x+3)}+\\dfrac{(x-3)(x+3)}{(x-2)(x+3)}-\\dfrac{(x-2)(x-2)}{(x-2)(x+3)} \\right] \\\\ & \\,\\,\\,\\,=\\dfrac{{{x}^{2}}-3x-{{x}^{2}}+9}{(x-3)(x+3)}:\\dfrac{9-{{x}^{2}}+({{x}^{2}}-9)-({{x}^{2}}-4x+4)}{(x-2)(x+3)} \\\\ & \\,\\,\\,\\,=\\dfrac{-3(x-3)}{(x-3)(x+3)}:\\dfrac{9-{{x}^{2}}+{{x}^{2}}-9-{{x}^{2}}+4x-4}{(x-2)(x+3)} \\\\ & \\,\\,\\,\\,=\\dfrac{-3}{x+3}:\\dfrac{-{{x}^{2}}+4x-4}{(x-2)(x+3)} \\\\ & \\,\\,\\,\\,=\\dfrac{-3}{x+3}.\\dfrac{(x-2)(x+3)}{-{{(x-2)}^{2}}} \\\\ & \\,\\,\\,\\,=\\dfrac{3}{x-2} \\\\ \\end{aligned}$<\/span>"}],"id_ques":271},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $C=\\left( \\dfrac{{{x}^{2}}-3x}{{{x}^{2}}-9}-1 \\right):\\left( \\dfrac{9-{{x}^{2}}}{{{x}^{2}}+x-6}-\\dfrac{x-3}{2-x}-\\dfrac{x-2}{x+3} \\right)$<br\/><b> C\u00e2u b<\/b>: T\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a $C$, bi\u1ebft $|x+3|=2x-1$<br\/><\/span> ","select":["A. $C=\\dfrac{3}{2}$","B. $C=\\dfrac{3}{-6}$","C. $C=\\dfrac{3}{2}$ ho\u1eb7c $C=\\dfrac{-9}{8}$","D. $C=\\dfrac{3}{-6}$ ho\u1eb7c $C=\\dfrac{-9}{8}$"],"hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne \\{-3;2;3\\}$<br\/>Theo c\u00e2u a, ta c\u00f3: $C=\\dfrac{3}{x-2}$<br\/>Ta c\u00f3: <br\/>$|x+3|=2x-1\\,\\,\\left(\\text{\u0110i\u1ec1u ki\u1ec7n c\u00f3 nghi\u1ec7m: }\\,2x-1\\ge 0\\Leftrightarrow x\\ge \\dfrac{1}{2}\\right)\\\\ \\Leftrightarrow \\left[\\begin{aligned} & x+3=2x-1\\, \\\\ &x+3=-2x+1 \\,\\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left[\\begin{aligned} & x=4\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & x=\\dfrac{-2}{3} \\,(\\text{lo\u1ea1i})\\\\ \\end{aligned}\\right.$<br\/>V\u1edbi $x=4$, ta c\u00f3 $C=\\dfrac{3}{4-2}=\\dfrac{3}{2}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}],"id_ques":272},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang ho\u1eb7c sai","title_trans":"\u0110\u1ec1 chung cho hai c\u00e2u","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"<span class='basic_left'>Cho bi\u1ec3u th\u1ee9c $D=\\dfrac{x-1}{2}:\\left( \\dfrac{{{x}^{2}}+2}{{{x}^{3}}-1}+\\dfrac{x}{{{x}^{2}}+x+1}+\\dfrac{1}{1-x} \\right)$<br\/><b> C\u00e2u a<\/b>: $D<0$ v\u1edbi $x\\ne 1$. <b>\u0110\u00fang<\/b> hay <b>Sai<\/b>?<\/span>","select":[" \u0110\u00fang "," Sai "],"hint":"R\u00fat g\u1ecdn $D$ r\u1ed3i so s\u00e1nh v\u1edbi $0$","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 1$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & D=\\dfrac{x-1}{2}:\\left( \\dfrac{{{x}^{2}}+2}{{{x}^{3}}-1}+\\dfrac{x}{{{x}^{2}}+x+1}+\\dfrac{1}{1-x} \\right) \\\\ & \\,\\,\\,\\,=\\dfrac{x-1}{2}:\\left[ \\dfrac{{{x}^{2}}+2}{(x-1)({{x}^{2}}+x+1)}+\\dfrac{x}{{{x}^{2}}+x+1}-\\dfrac{1}{x-1} \\right] \\\\ & \\,\\,\\,\\,=\\dfrac{x-1}{2}:\\left[ \\dfrac{{{x}^{2}}+2}{(x-1)({{x}^{2}}+x+1)}+\\dfrac{x(x-1)}{(x-1)({{x}^{2}}+x+1)}-\\dfrac{{{x}^{2}}+x+1}{(x-1)({{x}^{2}}+x+1)} \\right] \\\\ & \\,\\,\\,\\,=\\dfrac{x-1}{2}:\\dfrac{{{x}^{2}}+2+{{x}^{2}}-x-({{x}^{2}}+x+1)}{(x-1)({{x}^{2}}+x+1)} \\\\ & \\,\\,\\,\\,=\\dfrac{x-1}{2}:\\dfrac{{{x}^{2}}-2x+1}{(x-1)({{x}^{2}}+x+1)} \\\\ & \\,\\,\\,\\,=\\dfrac{x-1}{2}.\\dfrac{(x-1)({{x}^{2}}+x+1)}{{{(x-1)}^{2}}} \\\\ & \\,\\,\\,\\,=\\dfrac{{{x}^{2}}+x+1}{2} \\\\ \\end{aligned}$<br\/>Ta c\u00f3: $\\left\\{ \\begin{align} & {{x}^{2}}+x+1={{x}^{2}}+2.x.\\dfrac{1}{2}+\\dfrac{1}{4}+\\dfrac{3}{4}={{\\left( x+\\dfrac{1}{2} \\right)}^{2}}+\\dfrac{3}{4}>0\\,\\,\\forall \\,x \\\\ & 2>0 \\\\ \\end{align} \\right.$<br\/>Do \u0111\u00f3 $D > 0$ v\u1edbi m\u1ecdi $x\\ne 1$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 <b>Sai<\/b>.<\/span><\/span>","column":2}],"id_ques":273},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"],["8"],["-1"],["2"]]],"list":[{"point":5,"width":50,"ques":"Cho bi\u1ec3u th\u1ee9c $D=\\dfrac{x-1}{2}:\\left( \\dfrac{{{x}^{2}}+2}{{{x}^{3}}-1}+\\dfrac{x}{{{x}^{2}}+x+1}+\\dfrac{1}{1-x} \\right)$<br\/><b> C\u00e2u b<\/b>: T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a $D$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $Min\\, D=$<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div> khi $x=$<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"Bi\u1ebfn \u0111\u1ed5i bi\u1ec3u th\u1ee9c $D$ v\u1ec1 d\u1ea1ng $\\dfrac{f^2(x)+m}{n}$ v\u1edbi $m,n$ l\u00e0 c\u00e1c h\u1eb1ng s\u1ed1, r\u1ed3i \u0111\u00e1nh gi\u00e1.","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 1$<br\/>Theo c\u00e2u a, ta c\u00f3: $D=\\dfrac{x^2+x+1}{2}=\\dfrac{{{x}^{2}}+2.x.\\dfrac{1}{2}+\\dfrac{1}{4}+\\dfrac{3}{4}}{2}=\\dfrac{{{\\left( x+\\dfrac{1}{2} \\right)}^{2}}+\\dfrac{3}{4}}{2}$<br\/>Ta c\u00f3: <br\/>$\\left( x+\\dfrac{1}{2} \\right)^{2} \\ge 0 \\,\\forall \\,x \\Rightarrow \\left( x+\\dfrac{1}{2} \\right)^{2}+\\dfrac{3}{4} \\ge \\dfrac{3}{4} \\forall \\,x \\Rightarrow D \\ge \\dfrac{3}{8}\\, \\forall \\,x$<br\/>V\u1eady gi\u00e1 nh\u1ecf nh\u1ea5t c\u1ee7a $D$ l\u00e0 $\\dfrac{3}{8}$.<br\/>D\u1ea5u $\u201c=\u201d$ x\u1ea3y ra khi $\\left( x+\\dfrac{1}{2} \\right)^{2}=0 \\Leftrightarrow x=\\dfrac{-1}{2}$<\/span>"}],"id_ques":274},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang ho\u1eb7c sai","title_trans":"","temp":"true_false","correct":[["f","t","t","f","t"]],"list":[{"point":5,"image":"","col_name":[""," \u0110\u00fang ","<b>Sai<\/b>"],"arr_ques":["H\u00ecnh thang c\u00f3 hai c\u1ea1nh b\u00ean b\u1eb1ng nhau l\u00e0 h\u00ecnh thang c\u00e2n. ","T\u1ee9 gi\u00e1c c\u00f3 $4$ g\u00f3c b\u1eb1ng nhau l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt.","T\u1ee9 gi\u00e1c c\u00f3 hai \u0111\u01b0\u1eddng ch\u00e9o b\u1eb1ng nhau v\u00e0 vu\u00f4ng g\u00f3c v\u1edbi nhau l\u00e0 h\u00ecnh vu\u00f4ng.","L\u1ee5c gi\u00e1c \u0111\u1ec1u m\u1ed7i g\u00f3c b\u1eb1ng $60^o$","\u0110a gi\u00e1c l\u1ed3i c\u00f3 $8$ c\u1ea1nh th\u00ec c\u00f3 $20$ \u0111\u01b0\u1eddng ch\u00e9o"],"hint":"","explain":["<span class='basic_left'><b>Sai<\/b>. V\u00ec h\u00ecnh thang c\u00f3 hai c\u1ea1nh b\u00ean b\u1eb1ng nhau c\u00f3 th\u1ec3 l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh<\/span>","<br\/><span class='basic_left'><b>\u0110\u00fang<\/b>. V\u00ec t\u1ed5ng $4$ g\u00f3c trong m\u1ed9t t\u1ee9 gi\u00e1c b\u1eb1ng $360 ^o$ m\u00e0 $4$ g\u00f3c b\u1eb1ng nhau n\u00ean s\u1ed1 \u0111o m\u1ed7i g\u00f3c l\u00e0 $90 ^o$. Do v\u1eady t\u1ee9 gi\u00e1c l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt.<\/span>","<br\/><span class='basic_left'><b>Sai<\/b>. N\u1ebfu hai \u0111\u01b0\u1eddng ch\u00e9o kh\u00f4ng c\u1eaft nhau t\u1ea1i trung \u0111i\u1ec3m m\u1ed7i \u0111\u01b0\u1eddng. Nh\u01b0 h\u00ecnh sau:<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai24/lv3/img\/D8_B24_1.png' \/><\/center><\/span>","<br\/><span class='basic_left'><b>Sai<\/b>. V\u00ec l\u1ee5c gi\u00e1c \u0111\u1ec1u c\u00f3 s\u1ed1 \u0111o m\u1ed7i g\u00f3c l\u00e0 <br\/>$\\dfrac{(6-2).180^o}{6}=120^o\\ne 60^o$.<\/span> ","<br\/><span class='basic_left'><b>\u0110\u00fang<\/b>. V\u00ec s\u1ed1 \u0111\u01b0\u1eddng ch\u00e9o c\u1ee7a \u0111a gi\u00e1c $8$ c\u1ea1nh l\u00e0 $\\dfrac{(8-3).8}{2}=20$<\/span>"]}],"id_ques":275},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"ques":"Cho h\u00ecnh b\u00ecnh h\u00e0nh $ABCD$ c\u00f3 $AB=2BC$. G\u1ecdi $E$ v\u00e0 $F$ theo th\u1ee9 t\u1ef1 l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $CD$. G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $DE$ v\u00e0 $AF$, $ K$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $CE $ v\u00e0 $BF$. T\u1ee9 gi\u00e1c $EIFK$ l\u00e0 h\u00ecnh g\u00ec?","select":["A. H\u00ecnh b\u00ecnh h\u00e0nh","B. H\u00ecnh thoi","C. H\u00ecnh ch\u1eef nh\u1eadt","D. H\u00ecnh vu\u00f4ng"],"hint":"","explain":" <span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai24/lv3/img\/D8_B24_2.png' \/><\/center>$ABCD$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh n\u00ean ta c\u00f3: $AB\/\/DC; \\, AD\/\/BC$ v\u00e0 $AB=DC;\\,AD=BC$<br\/>V\u00ec $AB=2BC$ (gi\u1ea3 thi\u1ebft), suy ra $BC=AD=\\dfrac{AB}{2}=\\dfrac{DC}{2}$<br\/>L\u1ea1i c\u00f3 $AE=EB=\\dfrac{AB}{2}$; $DF=CF=\\dfrac{DC}{2}$ (gi\u1ea3 thi\u1ebft).<br\/> $\\Rightarrow AE=EB=BC=CF=FD=AD$<br\/>M\u1eb7t kh\u00e1c,$AD\/\/BC$ v\u00e0 $AE=EB;\\,DF=FC$, suy ra $EF\/\/AD\/\/BC$. <br\/>X\u00e9t t\u1ee9 gi\u00e1c $AEFD$ c\u00f3:<br\/> +) $AE\/\/DF;$<br\/>+) $ AD\/\/EF$ <br\/>$ \\Rightarrow ADFE$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)<br\/>L\u1ea1i c\u00f3 $AE=AD$ (ch\u1ee9ng minh tr\u00ean), n\u00ean $ADFE$ l\u00e0 h\u00ecnh thoi.<br\/>Do \u0111\u00f3, $AF\\bot DE \\Rightarrow \\widehat {EIF}=90^o$ (1).<br\/>Ch\u1ee9ng minh t\u01b0\u01a1ng t\u1ef1, ta c\u00f3: $BCFE$ l\u00e0 h\u00ecnh thoi suy ra $EC \\bot BF\\Rightarrow \\widehat{EKF}=90^o$ (2)<br\/>M\u1eb7t kh\u00e1c, $EF=DF=FC$ n\u00ean $\\Delta DEF$ vu\u00f4ng t\u1ea1i $E$ hay $\\widehat {DEF}=90^o$(3).<br\/>T\u1eeb (1), (2) v\u00e0 (3), $EIFK$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}],"id_ques":276},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang ho\u1eb7c sai","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"Cho tam gi\u00e1c $ABC$. G\u1ecdi $M, N, P$ l\u1ea7n l\u01b0\u1ee3t l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC, AC$ v\u00e0 $AB$. \u0110\u1ec3 $BCNP$ l\u00e0 h\u00ecnh thang c\u00e2n th\u00ec $ABC$ l\u00e0 tam gi\u00e1c c\u00e2n. <b> <b>\u0110\u00fang<\/b> <\/b> hay <b> <b>Sai<\/b><\/b>?","select":[" \u0110\u00fang "," Sai "],"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai24/lv3/img\/D8_B24_3.png' \/><\/center>X\u00e9t tam gi\u00e1c $ABC$ c\u00f3 $AP=PB$ (gi\u1ea3 thi\u1ebft) v\u00e0 $AN=NC$ (gi\u1ea3 thi\u1ebft)<br\/>Suy ra, $PN$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c $ABC$<br\/>Do v\u1eady, $PN\/\/BC$. Suy ra $BPNC$ l\u00e0 h\u00ecnh thang.<br\/>\u0110\u1ec3 $BCNP$ l\u00e0 h\u00ecnh thang c\u00e2n th\u00ec $\\widehat{B}=\\widehat{C}$ hay tam gi\u00e1c $ABC$ c\u00e2n t\u1ea1i $A$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang<\/span><\/span>","column":2}],"id_ques":277},{"time":3,"title":"Cho tam gi\u00e1c $ABC$ c\u00e2n \u1edf $A$. K\u1ebb $AH$ vu\u00f4ng g\u00f3c v\u1edbi $BC$ ($H \\in BC$). G\u1ecdi $M, N$ theo th\u1ee9 t\u1ef1 l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AB$ v\u00e0 $AC$. G\u1ecdi $E$ l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $H$ qua $M$. Ch\u1ee9ng minh $AH, MN$ v\u00e0 $EC$ \u0111\u1ed3ng quy","title_trans":"S\u1eafp x\u1ebfp c\u00e1c b\u01b0\u1edbc sau \u0111\u1ec3 ho\u00e0n th\u00e0nh b\u00e0i to\u00e1n","temp":"sequence","correct":[[[2],[5],[4],[1],[6],[3]]],"list":[{"point":5,"image":"","left":["Ta l\u1ea1i c\u00f3 $E$ l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $H$ qua $M$ n\u00ean $MH=\\dfrac{1}{2}HE$<br\/>Do \u0111\u00f3, $HE=AC$","M\u1eb7t kh\u00e1c, ta c\u00f3: $EM=MH;\\,EI=IC$ n\u00ean $MI\/\/BC$ (1).<br\/>T\u01b0\u01a1ng t\u1ef1 ta c\u00f3: $AM=MB;\\, AN=NC$ n\u00ean $MN\/\/BC$(2). <br\/>T\u1eeb (1) v\u00e0 (2), suy ra $M,I,N$ th\u1eb3ng h\u00e0ng. (**)","G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $EC$ (*). Khi \u0111\u00f3, $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $EC$","X\u00e9t tam gi\u00e1c $ABC$ c\u00f3 $MB=MA$ v\u00e0 $HB=HC$ n\u00ean $MH$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c $ABC$<br\/>Suy ra, $MH\/\/AC; \\,MH=\\dfrac{1}{2}AC$","T\u1eeb (*) v\u00e0 (**). Suy ra, $EC,MN,AH$ \u0111\u1ed3ng quy.","X\u00e9t t\u1ee9 gi\u00e1c $ACHE$ c\u00f3 $HE\/\/AC; HE=AC$, suy ra $ACHE$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh."],"top":100,"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai24/lv3/img\/D8_B24_4.png' \/><\/center><br\/>$ABC$ c\u00e2n t\u1ea1i $A$ c\u00f3 $AH$ l\u00e0 \u0111\u01b0\u1eddng cao n\u00ean $H$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $BC$. (t\u00ednh ch\u1ea5t tam gi\u00e1c c\u00e2n)<br\/>X\u00e9t tam gi\u00e1c $ABC$ c\u00f3:<br\/>+) $MB=MA$ (gi\u1ea3 thi\u1ebft) <br\/> +) $HB=HC$ (ch\u1ee9ng minh tr\u00ean) <br\/>$ \\Rightarrow MH$ l\u00e0 \u0111\u01b0\u1eddng trung b\u00ecnh c\u1ee7a tam gi\u00e1c $ABC$.<br\/> Suy ra, $MH\/\/AC; \\,MH=\\dfrac{1}{2}AC$<br\/>Ta l\u1ea1i c\u00f3 $E$ l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $H$ qua $M$ n\u00ean $MH=\\dfrac{1}{2}HE$<br\/>Do \u0111\u00f3, $HE=AC$<br\/>X\u00e9t t\u1ee9 gi\u00e1c $ACHE$ c\u00f3:<br\/>+) $HE\/\/AC$; <br\/>+) $HE=AC$<br\/>Suy ra $ACHE$ l\u00e0 h\u00ecnh b\u00ecnh h\u00e0nh. (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)<br\/>G\u1ecdi $I$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $EC$ (*). Khi \u0111\u00f3, $I$ l\u00e0 trung \u0111i\u1ec3m c\u1ee7a $AH$ v\u00e0 $EC$ (t\u00ednh ch\u1ea5t h\u00ecnh b\u00ecnh h\u00e0nh)<br\/>M\u1eb7t kh\u00e1c, ta c\u00f3: $EM=MH;\\,EI=IC$ (ch\u1ee9ng minh tr\u00ean)<br\/> N\u00ean $MI\/\/BC$ (1).<br\/>T\u01b0\u01a1ng t\u1ef1 ta c\u00f3: $AM=MB;\\, AN=NC$ (gi\u1ea3 thi\u1ebft)<br\/>N\u00ean $MN\/\/BC$(2). <br\/>T\u1eeb (1) v\u00e0 (2), suy ra $M,I,N$ th\u1eb3ng h\u00e0ng. (**). <br\/>T\u1eeb (*) v\u00e0 (**). Suy ra, $EC,MN,AH$ \u0111\u1ed3ng quy.<\/span>"}],"id_ques":278},{"time":3,"title":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ v\u00e0 $M$ l\u00e0 trung \u0111i\u1ec3m c\u1ea1nh $BC$. T\u1eeb $M$ k\u1ebb $MD$ vu\u00f4ng g\u00f3c v\u1edbi $AB$ t\u1ea1i $D$ v\u00e0 $ME$ vu\u00f4ng g\u00f3c v\u1edbi $AC$ t\u1ea1i $E$. G\u1ecdi $P$ l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng c\u1ee7a $D$ qua $M$; $Q$ l\u00e0 \u0111i\u1ec3m \u0111\u1ed1i x\u1ee9ng v\u1edbi $E$ qua $M$. <br\/>T\u00ednh t\u1ec9 s\u1ed1 $\\dfrac{PQ}{BC}$","title_trans":"S\u1eafp x\u1ebfp c\u00e1c b\u01b0\u1edbc sau \u0111\u1ec3 ho\u00e0n th\u00e0nh b\u00e0i to\u00e1n","temp":"sequence","correct":[[[2],[4],[3],[1]]],"list":[{"point":5,"image":"","left":["M\u1eb7t kh\u00e1c, $\\Delta ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $M$ l\u00e0 trung \u0111i\u1ec3m $BC$ n\u00ean $AM=BM=MC$<br\/>Do v\u1eady, $DE=AM=\\dfrac{BC}{2}$. "," V\u1eady $QP=DE=\\dfrac{BC}{2}\\Rightarrow \\dfrac{PQ}{BC}=\\dfrac{1}{2}$","X\u00e9t $\\Delta MDE$ v\u00e0 $\\Delta MPQ$ c\u00f3: $\\widehat {DME}=\\widehat {QMP}$ (\u0111\u1ed1i \u0111\u1ec9nh); $MD=MP$ (gi\u1ea3 thi\u1ebft); $MQ=ME$ (gi\u1ea3 thi\u1ebft)<br\/>Do \u0111\u00f3, $\\Delta MDE=\\Delta MPQ$ (c.g.c). Suy ra, $ QP=DE=\\dfrac{BC}{2}$","X\u00e9t t\u1ee9 gi\u00e1c $AEMD$ c\u00f3 $\\widehat{A}=90^o$ (gi\u1ea3 thi\u1ebft); $\\widehat {ADM}=90^o$ (v\u00ec $MD\\bot AB$); $\\widehat {AEM}=90^o$ (v\u00ec $ME\\bot AC$). Do \u0111\u00f3, $AEMD$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt. Suy ra, $AM=DE$ (t\u00ednh ch\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt)"],"top":105,"hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai24/lv3/img\/D8_B24_5.png' \/><\/center><br\/>X\u00e9t t\u1ee9 gi\u00e1c $AEMD$ c\u00f3:<br\/>+) $\\widehat{A}=90^o$ (gi\u1ea3 thi\u1ebft);<br\/>+) $\\widehat {ADM}=90^o$ (v\u00ec $MD\\bot AB$);<br\/>+) $\\widehat {AEM}=90^o$ (v\u00ec $ME\\bot AC$)<br\/>$\\Rightarrow AEMD$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt (d\u1ea5u hi\u1ec7u nh\u1eadn bi\u1ebft)<br\/>Suy ra, $AM=DE$ (t\u00ednh ch\u1ea5t h\u00ecnh ch\u1eef nh\u1eadt)<br\/>M\u1eb7t kh\u00e1c, $\\Delta ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3 $M$ l\u00e0 trung \u0111i\u1ec3m $BC$<br\/> N\u00ean $AM=BM=MC$<br\/>Do v\u1eady, $DE=AM=\\dfrac{BC}{2}$<br\/>X\u00e9t $\\Delta MDE$ v\u00e0 $\\Delta MPQ$ c\u00f3:<br\/>+) $MD=MP$ (gi\u1ea3 thi\u1ebft)<br\/>+) $\\widehat {DME}=\\widehat {QMP}$ (\u0111\u1ed1i \u0111\u1ec9nh)<br\/>+) $MQ=ME$ (gi\u1ea3 thi\u1ebft)<br\/>Suy ra $\\Delta MDE=\\Delta MPQ$ (c.g.c)<br\/>$\\Rightarrow QP=DE=\\dfrac{BC}{2}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng)<br\/>V\u1eady $\\dfrac{PQ}{BC}=\\dfrac{1}{2}$<\/span>"}],"id_ques":279},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["2"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"Cho $x=by+cz$; $y=ax+cz$; $z=ax+by$ v\u00e0 $x+y+x \\ne 0$. <br\/>T\u00ednh $\\dfrac{1}{1+a}+\\dfrac{1}{1+b}+\\dfrac{1}{1+c}=$_input_","hint":"","explain":"<span class='basic_left'><br\/>Ta c\u00f3: $x=by+cz\\Leftrightarrow ax+x=ax+by+cz \\Leftrightarrow (1+a)x=ax+by+cz\\Leftrightarrow 1+a=\\dfrac{ax+by+cz}{x}$<br\/>T\u01b0\u01a1ng t\u1ef1, ta c\u00f3:<br\/>$y=ax+cz\\Leftrightarrow by+y=ax+by+cz \\Leftrightarrow (1+b)y=ax+by+cz\\Leftrightarrow 1+b=\\dfrac{ax+by+cz}{y}$<br\/>$z=ax+by\\Leftrightarrow cz+z=ax+by+cz \\Leftrightarrow (1+c)z=ax+by+cz\\Leftrightarrow 1+c=\\dfrac{ax+by+cz}{z}$<br\/>Do v\u1eady, ta c\u00f3:<br\/>$\\dfrac{1}{1+a}+\\dfrac{1}{1+b}+\\dfrac{1}{1+c}\\\\ =\\dfrac{1}{\\dfrac{ax+by+cz}{x}}+\\dfrac{1}{\\dfrac{ax+by+cz}{y}}+\\dfrac{1}{\\dfrac{ax+by+cz}{z}}\\\\ =\\dfrac{x}{ax+by+cz}+\\dfrac{y}{ax+by+cz}+\\dfrac{z}{ax+by+cz}=\\dfrac{x+y+z}{ax+by+cz}$<br\/>M\u00e0 ta l\u1ea1i c\u00f3: $x+y+z=by+cz+ax+cz+ax+by=2(ax+by+cz)$<br\/>Do v\u1eady, $\\dfrac{x+y+z}{ax+by+cz}=\\dfrac{2(ax+by+cz)}{ax+by+cz}=2$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$<\/span><\/span>"}],"id_ques":280}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":59}}