{"segment":[{"part":[{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["2","3","4"]],"list":[{"point":5,"img":"","ques":"Trong c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau ph\u01b0\u01a1ng tr\u00ecnh n\u00e0o \u0111\u01b0a \u0111\u01b0\u1ee3c v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n?","hint":"","column":1,"number_true":2,"select":["A. $2(x-1)+2(3-x)=3$","B. $3(x^2-x+1)-3x^2+5=0$","C. $x+\\dfrac{1}{x-1}=x+1$","D. $\\dfrac{1}{x-2}+\\dfrac{1}{x}=\\dfrac{-2}{x(x-2)}$"],"explain":"<span class='basic_left'>A. $2(x-1)+2(3-x)=3\\\\ \\Leftrightarrow 2x-2+6-2x=3\\\\ \\Leftrightarrow 0.x=-1$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh A kh\u00f4ng l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n.<br\/>B. $3(x^2-x+1)-3x^2+5=0\\\\ \\Leftrightarrow 3x^2-3x+3-3x^2+5=0\\\\ \\Leftrightarrow -3x+8=0$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh B l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n<br\/>C. \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 1$.<br\/>$x+\\dfrac{1}{x-1}=x+1\\\\ \\Rightarrow x(x-1)+1=(x+1)(x-1)\\\\ \\Leftrightarrow x^2-x+1-(x^2-1)=0\\\\ \\Leftrightarrow -x+2=0$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh C l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n.<br\/>D. \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 2$<br\/> $\\dfrac{1}{x-2}+\\dfrac{1}{x}=\\dfrac{-2}{x(x-2)}\\\\ \\Leftrightarrow \\dfrac{x}{x(x-2)}+\\dfrac{x-2}{x(x-2)}=\\dfrac{-2}{x(x-2)}\\\\ \\Rightarrow x+x-2=-2\\\\ \\Leftrightarrow 2x=0$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh D l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n.<\/span>"}],"id_ques":1261},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["4"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $|x-3|=9-2x$.<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$","hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m $9-2x\\ge 0 \\Rightarrow x\\le \\dfrac{9}{2}$<br\/>V\u1edbi $x\\le \\dfrac{9}{2}$. Ta c\u00f3:<br\/>$|x-3|=9-2x\\\\ \\Leftrightarrow \\left [\\begin{aligned}&x-3=9-2x\\\\ &x-3=2x-9\\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left [\\begin{aligned}&3x=12\\\\ &x=6\\\\ \\end{aligned}\\right. \\\\ \\Leftrightarrow \\left [\\begin{aligned}&x=4\\\\ &x=6\\,\\,\\text{(lo\u1ea1i)}\\\\ \\end{aligned}\\right.$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{4\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $4$<\/span><\/span>"}],"id_ques":1262},{"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/bai30/lv3/img\/12.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $(3x-1)(x+2)=(3x-1)(7x-10)$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left \\{\\dfrac{-1}{3};\\dfrac{-12}{5}\\right\\}$","B. $S=\\left \\{\\dfrac{1}{3};\\dfrac{12}{5}\\right\\}$","C. $S=\\left \\{\\dfrac{1}{3};2\\right\\}$","D. $S=\\left \\{\\dfrac{-1}{3};2\\right\\}$"],"hint":"","explain":" <span class='basic_left'>$\\begin{aligned} & (3x-1)(x+2)=(3x-1)(7x-10) \\\\ & \\Leftrightarrow (3x-1)(x+2)-(3x-1)(7x-10)=0 \\\\ & \\Leftrightarrow (3x-1)[x+2-(7x-10)]=0 \\\\ & \\Leftrightarrow (3x-1)(-6x+12)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & 3x-1=0 \\\\ & -6x+12=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=\\dfrac{1}{3} \\\\ & x=2\\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left \\{\\dfrac{1}{3};2\\right\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}],"id_ques":1263},{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. {$\\dfrac{3}{2};3$}","B. {$\\dfrac{3}{5};2$}","C. {$\\dfrac{2}{5};2$}"],"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $2x(x-3)-3x+9=0$.<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=?$","hint":"","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$2x(x-3)-3x+9=0\\\\ \\Leftrightarrow 2x(x-3)-3(x-3)=0\\\\ \\Leftrightarrow (2x-3)(x-3)=0\\\\ \\Leftrightarrow \\left[\\begin{aligned}&2x-3=0\\\\ &x-3=0\\\\ \\end{aligned}\\right. \\\\ \\Leftrightarrow \\left[\\begin{aligned}& x=\\dfrac{3}{2} \\\\ & x=3 \\\\ \\end{aligned}\\right.$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{\\dfrac{3}{2};3\\right\\}$<\/span>"}],"id_ques":1264},{"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/bai30/lv3/img\/9.jpg' \/><\/center> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{2}{{{x}^{2}}+2x+1}-\\dfrac{5}{{{x}^{2}}-2x+1}=\\dfrac{3}{{{x}^{2}}-1}$.<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\left\\{0;\\dfrac{7}{3}\\right\\}$","B. $S=\\left\\{0;-\\dfrac{7}{3}\\right\\}$","C. $S=\\left\\{0;-\\dfrac{3}{7}\\right\\}$","D. $S=\\left\\{-\\dfrac{7}{3}\\right\\}$"],"hint":"","explain":" <span class='basic_left'><br\/>\u0110i\u1ec1u ki\u1ec7n: $x\\ne \\pm 1$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & \\dfrac{2}{{{x}^{2}}+2x+1}-\\dfrac{5}{{{x}^{2}}-2x+1}=\\dfrac{3}{{{x}^{2}}-1} \\\\ & \\Leftrightarrow \\dfrac{2}{{{(x+1)}^{2}}}-\\dfrac{5}{{{(x-1)}^{2}}}=\\dfrac{3}{(x-1)(x+1)} \\\\ & \\Leftrightarrow \\dfrac{2{{(x-1)}^{2}}}{{{(x-1)}^{2}}{{(x+1)}^{2}}}-\\dfrac{5{{(x+1)}^{2}}}{{{(x-1)}^{2}}{{(x+1)}^{2}}}=\\dfrac{3(x-1)(x+1)}{{{(x-1)}^{2}}{{(x+1)}^{2}}} \\\\ & \\Rightarrow 2{{(x-1)}^{2}}-5{{(x+1)}^{2}}=3({{x}^{2}}-1) \\\\ & \\Leftrightarrow 2({{x}^{2}}-2x+1)-5({{x}^{2}}+2x+1)-3({{x}^{2}}-1)=0 \\\\ & \\Leftrightarrow 2{{x}^{2}}-4x+2-5{{x}^{2}}-10x-5-3{{x}^{2}}+3=0 \\\\ & \\Leftrightarrow -6{{x}^{2}}-14x=0 \\\\ & \\Leftrightarrow 3{{x}^{2}}+7x=0 \\\\ & \\Leftrightarrow x(3x+7)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & 3x+7=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x=-\\dfrac{7}{3} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{0;-\\dfrac{7}{3}\\right\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}],"id_ques":1265},{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":5,"ques":"$S=\\{0;1\\}$ l\u00e0 t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $ \\dfrac{2x-1}{x-1}+\\dfrac{x}{(x-1)(x-2)}=\\dfrac{6x-2}{x-2}$","select":["\u0110\u00fang","Sai"],"hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n: $x\\ne 1;\\, x\\ne 2$<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & \\dfrac{2x-1}{x-1}+\\dfrac{x}{(x-1)(x-2)}=\\dfrac{6x-2}{x-2} \\\\ & \\Leftrightarrow \\dfrac{(2x-1)(x-2)}{(x-1)(x-2)}+\\dfrac{x}{(x-1)(x-2)}=\\dfrac{(6x-2)(x-1)}{(x-1)(x-2)} \\\\ & \\Rightarrow (2x-1)(x-2)+x=(6x-2)(x-1) \\\\ & \\Leftrightarrow 2{{x}^{2}}-4x-x+2+x-(6{{x}^{2}}-6x-2x+2)=0 \\\\ & \\Leftrightarrow 2{{x}^{2}}-4x+2-6{{x}^{2}}+8x-2=0 \\\\ & \\Leftrightarrow -4{{x}^{2}}+4x=0 \\\\ & \\Leftrightarrow {{x}^{2}}-x=0 \\\\ & \\Leftrightarrow x(x-1)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x-1=0 \\\\ \\end{aligned} \\right. \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0 \\\\ & x=1\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{0\\}$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai.<\/span><\/span>","column":2}],"id_ques":1266},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["40"],["60"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>L\u00fac $6$ gi\u1edd s\u00e1ng, m\u1ed9t xe m\u00e1y kh\u1edfi h\u00e0nh t\u1eeb $A$ \u0111\u1ec3 \u0111i \u0111\u1ebfn $B$. \u0110\u1ebfn $7$ gi\u1edd $30$ ph\u00fat m\u1ed9t \u00f4 t\u00f4 c\u0169ng kh\u1edfi h\u00e0nh t\u1eeb $A$ \u0111\u1ec3 \u0111i \u0111\u1ebfn $B$ v\u1edbi v\u1eadn t\u1ed1c l\u1edbn h\u01a1n v\u1eadn t\u1ed1c xe m\u00e1y l\u00e0 $20km\/h$ v\u00e0 hai xe g\u1eb7p nhau l\u00fac $10$ gi\u1edd $30$ ph\u00fat. T\u00ednh v\u1eadn t\u1ed1c c\u1ee7a xe m\u00e1y v\u00e0 \u00f4 t\u00f4?<br\/> (xe m\u00e1y v\u00e0 \u00f4 t\u00f4 kh\u00f4ng b\u1ecb h\u01b0 h\u1ecfng hay d\u1eebng l\u1ea1i d\u1ecdc \u0111\u01b0\u1eddng)<br\/><b> \u0110\u00e1p s\u1ed1: <\/b>V\u1eadn t\u1ed1c xe m\u00e1y: _input_ ($km\/h$); V\u1eadn t\u1ed1c \u00f4 t\u00f4: _input_ ($km\/h$)<\/span>","hint":"","explain":"<span class='basic_left'>G\u1ecdi v\u1eadn t\u1ed1c c\u1ee7a xe m\u00e1y l\u00e0 $x\\, (km\/h,\\,x>0)$<br\/>V\u00ec v\u1eadn t\u1ed1c \u00f4 t\u00f4 l\u1edbn h\u01a1n v\u1eadn t\u1ed1c xe m\u00e1y l\u00e0 $20\\,km\/h$ n\u00ean v\u1eadn t\u1ed1c \u00f4 t\u00f4 l\u00e0 $x+20$ ($km\/h$)<br\/>Xe m\u00e1y \u0111i t\u1eeb $6$ gi\u1edd \u0111\u1ebfn $10$ gi\u1edd $30$ ph\u00fat g\u1eb7p \u00f4 t\u00f4 n\u00ean th\u1eddi gian xe m\u00e1y \u0111i \u0111\u1ebfn l\u00fac g\u1eb7p \u00f4 t\u00f4 l\u00e0 $4 $ gi\u1edd $30$ ph\u00fat ($=\\dfrac{9}{2}$ gi\u1edd)<br\/>Do v\u1eady, qu\u00e3ng \u0111\u01b0\u1eddng xe m\u00e1y \u0111i \u0111\u01b0\u1ee3c \u0111\u1ebfn khi g\u1eb7p \u00f4 t\u00f4 l\u00e0 $\\dfrac{9x}{2}$ ($km$)<br\/>\u00d4 t\u00f4 \u0111i t\u1eeb $7$ gi\u1edd $3$0 ph\u00fat \u0111\u1ebfn $10$ gi\u1edd $30$ ph\u00fat g\u1eb7p xe m\u00e1y n\u00ean th\u1eddi gian \u00f4 t\u00f4 \u0111i \u0111\u1ebfn l\u00fac g\u1eb7p xe m\u00e1y l\u00e0 $3$ gi\u1edd.<br\/>Suy ra qu\u00e3ng \u0111\u01b0\u1eddng \u00f4 t\u00f4 \u0111i \u0111\u01b0\u1ee3c \u0111\u1ebfn khi g\u1eb7p xe m\u00e1y l\u00e0 $3(x+20)$ ($km$)<br\/>V\u00ec hai xe c\u00f9ng xu\u1ea5t ph\u00e1t t\u1eeb $A$ v\u00e0 g\u1eb7p nhau, n\u00ean qu\u00e3ng \u0111\u01b0\u1eddng hai xe \u0111i \u0111\u01b0\u1ee3c l\u00e0 nh\u01b0 nhau.<br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$3(x+20)=\\dfrac{9x}{2}\\\\ \\Leftrightarrow 6(x+20)=9x\\\\ \\Leftrightarrow 6x+120=9x\\\\ \\Leftrightarrow 3x=120\\\\ \\Leftrightarrow x=40\\,\\,\\text{(th\u1ecfa m\u00e3n)}$<br\/>V\u1eady v\u1eadn t\u1ed1c c\u1ee7a xe m\u00e1y l\u00e0 $40\\,km\/h$ v\u00e0 v\u1eadn t\u1ed1c c\u1ee7a \u00f4 t\u00f4 l\u00e0 $60\\,km\/h$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $40;60$<\/span><\/span>"}],"id_ques":1267},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"fill_the_blank","correct":[[["42"],["38"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/8.jpg' \/><\/center><span class='basic_left'> L\u1edbp $8A$ v\u00e0 $8B$ c\u00f3 $80$ h\u1ecdc sinh. Trong \u0111\u1ee3t g\u00f3p s\u00e1ch \u1ee7ng h\u1ed9 m\u1ed7i em l\u1edbp $8A$ g\u00f3p $2$ quy\u1ec3n v\u00e0 m\u1ed7i em l\u1edbp $8B$ g\u00f3p $3$ quy\u1ec3n n\u00ean c\u1ea3 hai l\u1edbp g\u00f3p \u0111\u01b0\u1ee3c $198$ quy\u1ec3n. T\u00ecm s\u1ed1 h\u1ecdc sinh c\u1ee7a m\u1ed7i l\u1edbp.<br\/>S\u1ed1 h\u1ecdc sinh l\u1edbp $8A$ v\u00e0 $8B$ l\u1ea7n l\u01b0\u1ee3t l\u00e0:_input_ (h\u1ecdc sinh) v\u00e0 _input_ (h\u1ecdc sinh)<\/span>","hint":"","explain":" <span class='basic_left'>G\u1ecdi s\u1ed1 h\u1ecdc sinh l\u1edbp $8A$ l\u00e0 $x\\,$(h\u1ecdc sinh, $x \\in\\mathbb N^*$)<br\/>V\u00ec t\u1ed5ng s\u1ed1 h\u1ecdc sinh hai l\u1edbp $8A$ v\u00e0 $8B$ l\u00e0 $80$ h\u1ecdc sinh n\u00ean s\u1ed1 h\u1ecdc sinh l\u1edbp $8B$ l\u00e0 $80-x$ (h\u1ecdc sinh)<br\/>M\u1ed7i h\u1ecdc sinh l\u1edbp $8A$ \u0111\u00f3ng g\u00f3p $2$ quy\u1ec3n s\u00e1ch n\u00ean s\u1ed1 s\u00e1ch l\u1edbp $8A$ \u0111\u00f3ng g\u00f3p l\u00e0 $2x$ (quy\u1ec3n)<br\/>M\u1ed7i h\u1ecdc sinh l\u1edbp $8B$ \u0111\u00f3ng g\u00f3p $3$ quy\u1ec3n s\u00e1ch n\u00ean s\u1ed1 s\u00e1ch l\u1edbp $8B$ \u0111\u00f3ng g\u00f3p l\u00e0 $3(80-x)$ (quy\u1ec3n).<br\/>V\u00ec c\u1ea3 hai l\u1edbp \u0111\u00f3ng g\u00f3p \u0111\u01b0\u1ee3c $198$ quy\u1ec3n n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$2x+3(80-x)=198\\\\ \\Leftrightarrow 2x+240-3x=198\\\\ \\Leftrightarrow x=42\\,\\text{(th\u1ecfa m\u00e3n)}$<br\/>V\u1eady l\u1edbp $8A$ c\u00f3 $42$ h\u1ecdc sinh v\u00e0 $8B$ c\u00f3 $80-42=38$ h\u1ecdc hinh.<br\/><span class='basic_pink'>V\u1eady s\u1ed1 ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $42;38$<\/span><\/span>"}],"id_ques":1268},{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","t","f","t"]],"list":[{"point":5,"image":"https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/5.jpg","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["<span class='basic_left'>a. $0x+7y<0$ l\u00e0 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n.<\/span>","<span class='basic_left'>b. $\\dfrac{-1}{2}x+4\\ge 0$ l\u00e0 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n.<\/span>","<span class='basic_left'>c. N\u1ebfu $x>y$ th\u00ec $-3x>-3y$<\/span>","<span class='basic_left'>d. N\u1ebfu $x\\ge y$ th\u00ec $x+(-3)\\ge y+(-3)$<\/span>"],"hint":"","explain":["a. \u0110\u00fang v\u00ec. $0x+7y<0 \\Leftrightarrow 7y<0$ l\u00e0 b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t \u1ea9n $y$","<br\/>b. \u0110\u00fang theo \u0111\u1ecbnh ngh\u0129a.","<br\/>c. Sai v\u00ec $x>y\\Leftrightarrow -3x<-3y$","<br\/>d. \u0110\u00fang."]}],"id_ques":1269},{"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/bai30/lv3/img\/12.jpg' \/><\/center>Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh $-x-2>21x-24$","select":["A. $x<1$","B. $x>1$","C. $x<-1$","D. $x>-1$"],"hint":"","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$-x-2>21x-24\\\\ \\Leftrightarrow 21x+x<24-2\\\\ \\Leftrightarrow 22x<22\\\\ \\Leftrightarrow x<1$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}],"id_ques":1270},{"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/bai30/lv3/img\/9.jpg' \/><\/center>Tr\u1ee5c s\u1ed1 n\u00e0o sau \u0111\u00e2y bi\u1ec3u di\u1ec5n t\u1eadp nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{-2x+5}{12}\\ge \\dfrac{5x+3}{3}-\\dfrac{8x+1}{4}$","select":["A. <img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/D8_B30_1.png' \/>","B. <img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/D8_B30_2.png' \/>","C. <img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/D8_B30_3.png' \/>","D. <img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/D8_B30_4.png' \/>"],"hint":"","explain":" <span class='basic_left'>$\\begin{aligned} & \\dfrac{-2x+5}{12}\\ge \\dfrac{5x+3}{3}-\\dfrac{8x+1}{4} \\\\ & \\Leftrightarrow \\dfrac{-2x+5}{12}\\ge \\dfrac{4(5x+3)}{12}-\\dfrac{3(8x+1)}{12} \\\\ & \\Leftrightarrow -2x+5\\ge 20x+12-(24x+3) \\\\ & \\Leftrightarrow -2x-20x+24x\\ge 12-3-5 \\\\ & \\Leftrightarrow 2x\\ge 4 \\\\ & \\Leftrightarrow x\\ge 2 \\\\ \\end{aligned}$<br\/>V\u1eady $x\\ge 2$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}],"id_ques":1271},{"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/bai30/lv3/img\/9.jpg' \/><\/center>Gi\u1ea3i b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh $x^2-5x+6 < 0$.","select":["A. $2 < x < 3$","B. $x<2$ ho\u1eb7c $x>3$","C. $x<2$ v\u00e0 $x>3$","D. M\u1ecdi $x$ thu\u1ed9c $R$"],"hint":"","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$x^2-5x+6 < 0\\\\ \\Leftrightarrow x^2-2x-3x+6 < 0\\\\ \\Leftrightarrow x(x-2)-3(x-2)<0\\\\ \\Leftrightarrow (x-2)(x-3) < 0$<br\/>Tr\u01b0\u1eddng h\u1ee3p 1: <br\/>$\\left\\{\\begin{aligned}&x-2>0\\\\ &x-3<0\\\\ \\end{aligned}\\right. \\Leftrightarrow \\left\\{\\begin{aligned}& x > 2\\\\ & x < 3\\\\ \\end{aligned}\\right. \\Leftrightarrow 2 < x < 3$ <br\/> Tr\u01b0\u1eddng h\u1ee3p 2:<br\/>$\\left\\{\\begin{aligned}&x-2<0\\\\ &x-3>0\\\\ \\end{aligned}\\right. \\Leftrightarrow \\left\\{\\begin{aligned}& x < 2 \\\\ & x > 3\\\\ \\end{aligned}\\right.$ (v\u00f4 l\u00ed)<br\/>V\u1eady $2 < x < 3$ l\u00e0 nghi\u1ec7m c\u1ee7a b\u1ea5t ph\u01b0\u01a1ng tr\u00ecnh.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}],"id_ques":1272},{"title":"\u0110i\u1ec1n d\u1ea5u $<;>;=$ th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["<"]]],"list":[{"point":5,"width":50,"type_input":"","ques":" Cho $m>n$. So s\u00e1nh $-8m+1$ v\u00e0 $-8n+1$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $-8m+1$_input_ $-8n+1$","hint":"","explain":"<span class='basic_left'>Ta c\u00f3: $m > n \\\\ \\Leftrightarrow -8m < -8n\\\\ \\Leftrightarrow -8m+1 < -8n+1$<br\/><span class='basic_pink'>V\u1eady d\u1ea5u ph\u1ea3i \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $<$<\/span><\/span>"}],"id_ques":1273},{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0110\u00fang hay Sai?","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"V\u1edbi m\u1ecdi $x;\\,y \\in \\mathbb R$ th\u00ec $x^2+y^2+6>4x+2y$","select":["\u0110\u00fang","Sai"],"hint":"","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$x^2+y^2+6>4x+2y\\\\ \\Leftrightarrow x^2-4x+y^2-2y+6>0\\\\ \\Leftrightarrow x^2-4x+4+y^2-2y+1+1>0\\\\ \\Leftrightarrow (x-2)^2+(y-1)^2+1>0 \\,(1)\\,$.<br\/>V\u00ec $(x-2)^2\\ge 0\\,\\forall \\,x; \\,(y-1)^2\\ge0\\forall\\,y$ n\u00ean (1) lu\u00f4n \u0111\u00fang $\\forall \\,x,y\\in \\mathbb R$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0110\u00fang.<\/span><\/span>","column":2}],"id_ques":1274},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["60"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"Cho h\u00ecnh l\u0103ng tr\u1ee5 \u0111\u1ee9ng tam gi\u00e1c c\u00f3 chi\u1ec1u cao $AA\u2019=6 cm$, \u0111\u00e1y l\u00e0 tam gi\u00e1c vu\u00f4ng c\u00f3 hai c\u1ea1nh g\u00f3c vu\u00f4ng $AB=5 cm$ v\u00e0 $AC= 4 cm$. T\u00ednh th\u1ec3 t\u00edch l\u0103ng tr\u1ee5 \u0111\u1ee9ng.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> _input_($cm^3$)","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/D8_B30_5.png' \/><\/center>Di\u1ec7n t\u00edch \u0111\u00e1y $ABC$ l\u00e0 $\\dfrac{1}{2}.AB.AC=\\dfrac{1}{2}.5.4=10\\,cm^2$<br\/>Th\u1ec3 t\u00edch h\u00ecnh l\u0103ng tr\u1ee5 l\u00e0 $AA'.S_{ABC}=6.10=60\\,cm^3$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $60$<\/span><\/span>"}],"id_ques":1275},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["16"],["81"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho h\u00ecnh thang vu\u00f4ng $ABCD$ c\u00f3 $AB\/\/CD$ $(\\widehat A=90^o)$, $AB=4cm$, $CD=9cm$, $AD =6 cm$. G\u1ecdi $O$ l\u00e0 giao \u0111i\u1ec3m c\u1ee7a $AC$ v\u00e0 $BD$. T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch hai tam gi\u00e1c $AOB$ v\u00e0 $COD$.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $\\dfrac{S_{AOB}}{S_{COD}}=\\dfrac{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}}$<\/span>","hint":"T\u1ec9 s\u1ed1 di\u1ec7n t\u00edch b\u1eb1ng b\u00ecnh ph\u01b0\u01a1ng t\u1ec9 s\u1ed1 \u0111\u1ed3ng d\u1ea1ng","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/D8_B30_6.png' \/><\/center><br\/>X\u00e9t tam gi\u00e1c $AOB$ v\u00e0 tam gi\u00e1c $COD$ c\u00f3:<br\/>+) $\\widehat {BOA}=\\widehat {DOC}$ (hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh)<br\/>+) $\\widehat {BAO}=\\widehat {DCO}$ (hai g\u00f3c so le trong)<br\/>Suy ra, $\\Delta AOB \\backsim \\Delta COD$ (g.g)<br\/>Khi \u0111\u00f3, ta c\u00f3: $\\dfrac{S_{AOB}}{S_{COD}}=\\left(\\dfrac{AB}{DC}\\right)^2=\\left(\\dfrac{4}{9}\\right)^2=\\dfrac{16}{81}$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $16;81$<\/span><\/span>"}],"id_ques":1276},{"time":3,"title":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, \u0111\u01b0\u1eddng cao $AH$ ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $ABC$ c\u1eaft $AH$ t\u1ea1i $I$, c\u1eaft $AC$ t\u1ea1i $D$. Ch\u1ee9ng minh $DA.DB=BI.DC$","title_trans":"S\u1eafp x\u1ebfp c\u00e1c \u00fd sau th\u00e0nh b\u00e0i to\u00e1n ch\u1ee9ng minh","temp":"sequence","correct":[[[3],[6],[1],[2],[5],[4]]],"list":[{"point":5,"image":"","left":["X\u00e9t $\\Delta ABD$ v\u00e0 $\\Delta HBI$ c\u00f3: +) $\\widehat {BAD}=\\widehat {IHB}=90^o$; +) $\\widehat {ABD}=\\widehat {HBI}$ (t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c). Suy ra, $\\Delta ABD \\backsim \\Delta HBI$ (g.g)","T\u1eeb (1), (2) v\u00e0 (3), suy ra: $\\dfrac{DB}{BI}=\\dfrac{DC}{DA} \\Rightarrow DB.DA=BI.DC$","X\u00e9t tam gi\u00e1c $AHB$ v\u00e0 tam gi\u00e1c $CAB$ c\u00f3: +) $\\widehat {AHB}=\\widehat {CAB}=90^o$; +) $\\widehat B$ chung. Suy ra, $\\Delta AHB \\backsim \\Delta CAB$ (g.g)","Ta c\u00f3, $\\dfrac{AB}{BC}=\\dfrac{HB}{AB}\\Rightarrow \\dfrac{BC}{AB}=\\dfrac{AB}{HB}$ (1)","M\u1eb7t kh\u00e1c, V\u00ec $BD$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $ABC$ n\u00ean $\\dfrac{DC}{DA}=\\dfrac{BC}{AB}$ (3)","Ta c\u00f3, $\\dfrac{BD}{BI}=\\dfrac{AB}{HB}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) "],"top":100,"hint":"Ch\u1ee9ng minh b\u1eb1ng ph\u01b0\u01a1ng ph\u00e1p b\u1eafc c\u1ea7u.","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/D8_B30_7.png' \/><\/center> X\u00e9t tam gi\u00e1c $AHB$ v\u00e0 tam gi\u00e1c $CAB$ c\u00f3:<br\/> +) $\\widehat {AHB}=\\widehat {CAB}=90^o$; <br\/>+) $\\widehat B$ chung. <br\/>Suy ra, $\\Delta AHB \\backsim \\Delta CAB$ (g.g)<br\/>Ta c\u00f3, $\\dfrac{AB}{BC}=\\dfrac{HB}{AB}\\Rightarrow \\dfrac{BC}{AB}=\\dfrac{AB}{HB}$ (1) <br\/>X\u00e9t $\\Delta ABD$ v\u00e0 $\\Delta HBI$ c\u00f3: <br\/>+) $\\widehat {BAD}=\\widehat {IHB}=90^o$;<br\/> +) $\\widehat {ABD}=\\widehat {HBI}$ (t\u00ednh ch\u1ea5t ph\u00e2n gi\u00e1c). <br\/>Suy ra, $\\Delta ABD \\backsim \\Delta HBI$ (g.g)<br\/>Ta c\u00f3, $\\dfrac{BD}{BI}=\\dfrac{AB}{HB}$ (c\u1eb7p c\u1ea1nh t\u01b0\u01a1ng \u1ee9ng) (2) <br\/>M\u1eb7t kh\u00e1c, V\u00ec $BD$ l\u00e0 tia ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $ABC$ n\u00ean $\\dfrac{DC}{DA}=\\dfrac{BC}{AB}$ (3) <br\/>T\u1eeb (1), (2) v\u00e0 (3), suy ra: $\\dfrac{DB}{BI}=\\dfrac{DC}{DA} \\Rightarrow DB.DA=BI.DC$<\/span>"}],"id_ques":1277},{"title":"Ch\u1ecdn <u>nh\u1eefng<\/u> kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang","title_trans":"Ch\u1ecdn \u0111\u01b0\u1ee3c nhi\u1ec1u \u0111\u00e1p \u00e1n","temp":"checkbox","correct":[["1","2","3","5"]],"list":[{"point":5,"img":"","ques":"Cho tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$, c\u00f3 $AB=6 cm$ v\u00e0 $BC=10cm$. \u0110\u01b0\u1eddng ph\u00e2n gi\u00e1c $BD$ ($D$ thu\u1ed9c $AC$). $DH$ vu\u00f4ng g\u00f3c $BC$ ($H$ thu\u1ed9c $BC$).","hint":"","column":1,"number_true":2,"select":["A. $AC=8\\,cm$","B. $\\Delta ABC \\backsim \\Delta HDC$","C. $AB.DC=HD.BC$","D. $\\dfrac{AD}{DC}=\\dfrac{AB}{AC}$","E. $BD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $ADH$"],"explain":"<span class='basic_left'><br\/><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/D8_B30_8.png' \/><\/center>A. \u0110\u00fang v\u00ec. X\u00e9t tam gi\u00e1c $ABC$ vu\u00f4ng t\u1ea1i $A$ c\u00f3: $AB=6\\,cm$ $BC=10\\,cm$.<br\/>\u00c1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3: $BC^2=AB^2+AC^2\\\\ \\Leftrightarrow AC=\\sqrt{BC^2-AB^2}\\\\ \\Leftrightarrow AC=\\sqrt{64}=8\\,(cm)$<br\/>B. \u0110\u00fang v\u00ec X\u00e9t $\\Delta ABC$ v\u00e0 $\\Delta HDC$ c\u00f3: <br\/>+) $\\widehat {A} =\\widehat {H} =90^o$<br\/>+) $\\widehat C$ chung.<br\/>N\u00ean $\\Delta ABC \\backsim \\Delta HDC$ (g.g)<br\/>C. \u0110\u00fang v\u00ec $\\Delta ABC \\backsim \\Delta HDC$, suy ra $\\dfrac {AB}{HD}=\\dfrac{BC}{DC} \\Rightarrow AB.DC=HD.BC$<br\/>D. Sai v\u00ec v\u1edbi $BD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $\\widehat {ABC}$. Theo t\u00ednh ch\u1ea5t \u0111\u01b0\u1eddng ph\u00e2n gi\u00e1c c\u1ee7a tam gi\u00e1c c\u00f3 $\\dfrac{AD}{DC}=\\dfrac{AB}{BC}$<br\/>E. \u0110\u00fang v\u00ec X\u00e9t $\\Delta ABD$ v\u00e0 $\\Delta HBD$ c\u00f3: <br\/>+) $\\widehat A =\\widehat H = 90^o$<br\/>+) $\\widehat {ABD}=\\widehat {HBD}$<br\/>+) $BD$ chung.<br\/>N\u00ean $\\Delta ABD = \\Delta HBD$ (c\u1ea1nh huy\u1ec1n - g\u00f3c nh\u1ecdn)<br\/>Suy ra, $\\widehat {ADB}=\\widehat {HDB}$. Suy ra, $BD$ l\u00e0 ph\u00e2n gi\u00e1c c\u1ee7a g\u00f3c $ADH$<\/span>"}],"id_ques":1278},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"K\u1ebft qu\u1ea3 \u0111\u1ec3 d\u01b0\u1edbi d\u1ea1ng ph\u00e2n s\u1ed1 t\u1ed1i gi\u1ea3n","temp":"fill_the_blank","correct":[[["9"],["50"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"frac","ques":"Cho h\u00ecnh ch\u1eef nh\u1eadt $ABCD$ c\u00f3 $AD=8 cm$; $AB=6 cm$ hai \u0111\u01b0\u1eddng ch\u00e9o $AC$ v\u00e0 $BD$ c\u1eaft nhau t\u1ea1i $O$. Qua $D$ k\u1ebb \u0111\u01b0\u1eddng th\u1eb3ng $d$ vu\u00f4ng g\u00f3c v\u1edbi $BD$, $d$ c\u1eaft $BC$ t\u1ea1i $E$. Qua $O$ k\u1ebb $OH$ vu\u00f4ng g\u00f3c v\u1edbi $DC$ ($H$ thu\u1ed9c $DC$). T\u00ednh t\u1ec9 s\u1ed1 di\u1ec7n t\u00edch tam gi\u00e1c $EHC$ v\u00e0 di\u1ec7n t\u00edch tam gi\u00e1c $EDB$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $\\dfrac{S_{EHC}}{S_{BDE}}=$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"","explain":"<span class='basic_left'><center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai30/lv3/img\/D8_B30_9.png' \/><\/center><br\/>V\u00ec $ABCD$ l\u00e0 h\u00ecnh ch\u1eef nh\u1eadt n\u00ean $\\widehat A =\\widehat B =\\widehat C =\\widehat D = 90^o$ v\u00e0 $AB=DC=6\\,cm; \\, AD=BC=8\\,cm$<br\/>Suy ra, $AC=BD=\\sqrt{AB^2+AD^2}=\\sqrt{100}=10\\,(cm)$<br\/>X\u00e9t $\\Delta CDE$ v\u00e0 $\\Delta DBE$ c\u00f3: <br\/>+) $\\widehat {BDE}=\\widehat {DCE}=90^o$<br\/>+) $\\widehat {DBE}=\\widehat {CDE}$ (c\u00f9ng ph\u1ee5 v\u1edbi $\\widehat {BDC}$)<br\/>Suy ra, $\\Delta CDE \\backsim \\Delta DBE$ (g.g)<br\/>Suy ra $\\dfrac{S_{CDE}}{S_{DBE}}=\\left(\\dfrac{CD}{DB}\\right)^2=\\dfrac{36}{100}=\\dfrac{9}{25}\\Rightarrow S_{CDE}=\\dfrac{9}{25}S_{DBE}$ <br\/>M\u00e0 ta l\u1ea1i c\u00f3, $\\Delta ODC$ c\u00e2n t\u1ea1i $O$ c\u00f3 $OH\\bot DC$, suy ra $H$ l\u00e0 trung \u0111i\u1ec3m $DC$.<br\/>Suy ra, $S_{HCE}=\\dfrac{1}{2}.S_{CDE}=\\dfrac{1}{2}.\\dfrac{9}{25}S_{BDE}=\\dfrac{9}{50}S_{BDE}$<\/span>"}],"id_ques":1279},{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["3"],["2"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"frac","ques":"Cho $x,y,z>0$ v\u00e0 $x+y+z\\le 3$<br\/>T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c $A=\\dfrac{1}{1+x}+\\dfrac{1}{1+y}+\\dfrac{1}{1+z}$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b>$Min\\,A=$ <div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>","hint":"\u0110\u1eb7t $1+x=a; 1+y=b;1+z=c$. S\u1eed d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c ph\u1ee5: $(a+b+c)\\left(\\dfrac{1}{a}+\\dfrac{1}{b}+\\dfrac{1}{c}\\right)\\ge 9$","explain":"<span class='basic_left'>\u0110\u1eb7t $1+x=a\\\\ 1+y=b\\\\1+z=c$<br\/>Khi \u0111\u00f3, $A=\\dfrac{1}{a}+\\dfrac{1}{b}+\\dfrac{1}{c}$<br\/>Ta c\u00f3: $a+b+c=3+x+y+z$. M\u00e0 $x+y+x\\le 3$ n\u00ean:<br\/>$a+b+c\\le 6 \\Rightarrow \\dfrac{1}{a+b+c}\\ge \\dfrac{1}{6}$<br\/>Ta s\u1ebd ch\u1ee9ng minh b\u1ea5t \u0111\u1eb3ng th\u1ee9c sau:<br\/>$(a+b+c)\\left(\\dfrac{1}{a}+\\dfrac{1}{b}+\\dfrac{1}{c}\\right)\\ge 9$<br\/>Th\u1eadt v\u1eady, ta c\u00f3:<br\/>$(a+b+c)\\left(\\dfrac{1}{a}+\\dfrac{1}{b}+\\dfrac{1}{c}\\right)\\\\=\\dfrac{a+b+c}{a}+\\dfrac{a+b+c}{b}+\\dfrac{a+b+c}{c}\\\\=1+\\dfrac{b}{a}+\\dfrac{c}{a}+\\dfrac{a}{b}+1+\\dfrac{c}{b}+\\dfrac{a}{c}+\\dfrac{b}{c}+1\\\\=3+\\left(\\dfrac{b}{a}+\\dfrac{a}{b}\\right)+\\left(\\dfrac{c}{a}+\\dfrac{a}{c}\\right)+\\left(\\dfrac{c}{b}+\\dfrac{b}{c}\\right)$<br\/>V\u1edbi $a,y,c >0 \\Leftrightarrow a,b,c>0$. \u00c1p d\u1ee5ng b\u1ea5t \u0111\u1eb3ng th\u1ee9c Cauchy. Ta c\u00f3: $\\dfrac{b}{a}+\\dfrac{a}{b}\\ge 2;\\,\\dfrac{c}{a}+\\dfrac{a}{c}\\ge 2;\\,\\dfrac{c}{b}+\\dfrac{b}{c}\\ge 2$<br\/>Do v\u1eady, $(a+b+c)\\left(\\dfrac{1}{a}+\\dfrac{1}{b}+\\dfrac{1}{c}\\right)\\ge 9\\Rightarrow \\dfrac{1}{a}+\\dfrac{1}{b}+\\dfrac{1}{c}\\ge \\dfrac{9}{a+b+c}\\ge \\dfrac{1}{6}.9= \\dfrac{3}{2}$<br\/>D\u1ea5u \"=\" x\u1ea3y ra khi $a=b=c \\Leftrightarrow x=y=z=1$<\/span>"}],"id_ques":1280}],"id_ques":0}],"lesson":{"save":1,"level":3,"time":59}}