{"segment":[{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00fang hay sai","title_trans":"","temp":"true_false","correct":[["t","f","f","f","t"]],"list":[{"point":5,"image":"","col_name":["","\u0110\u00fang","Sai"],"arr_ques":["$3$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $x^2-9=0$","$\\{5\\}$ l\u00e0 t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $x^2-25=0$","Ph\u01b0\u01a1ng tr\u00ecnh $x=x+1$ c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m.","Ph\u01b0\u01a1ng tr\u00ecnh $x+1=x^2-1$ c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $\\{-1\\}$","Ph\u01b0\u01a1ng tr\u00ecnh $(x-1)(x+1)=x^2-1$ c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $\\mathbb R$"],"explain":["<span class='basic_left'> \u0110\u00fang v\u00ec. $x^2-9=0\\Leftrightarrow (x-3)(x+3)=0\\Leftrightarrow \\left[\\begin{aligned}&x=3\\\\ &x=-3\\\\ \\end{aligned}\\right.$ <br\/><span class='basic_pink'>Do v\u1eady, $3$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<\/span><\/span>","<br\/> <span class='basic_left'>Sai v\u00ec. <br\/>$x^2-25=0\\Leftrightarrow (x-5)(x+5)=0\\Leftrightarrow \\left[\\begin{aligned}&x=5\\\\ &x=-5\\\\ \\end{aligned}\\right.$ <br\/><span class='basic_pink'>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-5;5\\}$<\/span><\/span>","<br\/> <span class='basic_left'>Sai v\u00ec. <br\/>Ph\u01b0\u01a1ng tr\u00ecnh n\u00e0y v\u00f4 nghi\u1ec7m do $x=x+1\\Leftrightarrow 0=1$ (v\u00f4 l\u00fd)<\/span>","<br\/> <span class='basic_left'>Sai v\u00ec<br\/> $x+1=x^2-1 \\\\ \\Leftrightarrow x+1=(x+1)(x-1) \\\\ \\Leftrightarrow (x+1)(x-2)=0 \\\\ \\Leftrightarrow \\left[\\begin{aligned}&x=-1\\\\ & x=2\\\\ \\end{aligned}\\right. $ <br\/><span class='basic_pink'>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $\\{-1;2\\}$<\/span><\/span>","<br\/><span class='basic_left'>\u0110\u00fang v\u00ec $(x-1)(x+1)=x^2-1$ lu\u00f4n \u0111\u00fang v\u1edbi m\u1ecdi $x$<\/span>"]}]}],"id_ques":1011},{"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/bai22/lv2/img\/12.jpg' \/><\/center>S\u1ed1 $\\dfrac {1}{3}$ l\u00e0 m\u1ed9t nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ","select":["A. $3x-1=1$","B. $9x^2-1=0$","C. $x^2-3x+2=0$","D. $-27x^3=1$"],"hint":"Gi\u1ea3i v\u00e0 t\u00ecm nghi\u1ec7m m\u1ed7i ph\u01b0\u01a1ng tr\u00ecnh.","explain":" <span class='basic_left'>$\\bullet \\,\\, 3x-1=1\\Leftrightarrow 3x=2\\Leftrightarrow x=\\dfrac {2}{3}$<br\/>$\\bullet \\,\\, 9x^2-1=0\\Leftrightarrow (3x-1)(3x+1)=0\\Leftrightarrow \\left[ \\begin {aligned} &x=\\dfrac {1}{3}\\\\ &x=\\dfrac {-1}{3}\\\\ \\end{aligned}\\right.$<br\/>V\u1eady $\\dfrac {1}{3}$ l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh B.<br\/>$\\bullet \\,\\,x^2-3x+2=0\\Leftrightarrow (x-1)(x-2)=0\\Leftrightarrow \\left[ \\begin {aligned} &x=1\\\\ &x=2\\\\ \\end{aligned}\\right.$<br\/>$\\bullet\\,\\, -27x^3=1\\Leftrightarrow x^3=\\dfrac {-1}{27}\\Leftrightarrow x=\\dfrac {-1}{3}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":1012},{"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/bai22/lv2/img\/2.jpg' \/><\/center>Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y <b>sai<\/b>?","select":["A. Ph\u01b0\u01a1ng tr\u00ecnh $3x-7=5-x$ c\u00f3 nghi\u1ec7m $x=3$","B. Ph\u01b0\u01a1ng tr\u00ecnh $4x+1=x-2$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=-1$","C. Ph\u01b0\u01a1ng tr\u00ecnh $2x+6 =-x$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=-2$","D. Ph\u01b0\u01a1ng tr\u00ecnh $2x+x=x-1$ c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=0$"],"hint":"Gi\u1ea3i m\u1ed7i ph\u01b0\u01a1ng tr\u00ecnh tr\u00ean \u0111\u1ec3 t\u00ecm \u0111\u00e1p \u00e1n.","explain":" <span class='basic_left'>Kh\u1eb3ng \u0111\u1ecbnh <b> Sai<\/b> l\u00e0 c\u00e2u D v\u00ec:<br\/>$2x+x=x-1\\Leftrightarrow 2x=-1\\Leftrightarrow x=\\dfrac {-1}{2}\\ne 0$<br\/>Kh\u1eb3ng \u0111\u1ecbnh A, B v\u00e0 C \u0111\u00fang v\u00ec<br\/>A. $3x-7=5-x\\Leftrightarrow 4x=12\\Leftrightarrow x=3$<br\/>B. $4x+1=x-2\\Leftrightarrow 3x=-3\\Leftrightarrow x=-1$<br\/>C. $2x+6=-x\\Leftrightarrow 3x=-6\\Leftrightarrow x=-2$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n l\u00e0 D<\/span><\/span>","column":1}]}],"id_ques":1013},{"time":24,"part":[{"time":3,"title":"N\u1ed1i m\u1ed7i ph\u01b0\u01a1ng tr\u00ecnh c\u1ed9t tr\u00e1i v\u1edbi t\u1eadp nghi\u1ec7m t\u01b0\u01a1ng \u1ee9ng \u1edf c\u1ed9t ph\u1ea3i. ","title_trans":"","audio":"","temp":"matching","correct":[["3","4","1","2"]],"list":[{"point":5,"image":"","left":["$7-(2x+4)=-(x+4)$ ","$\\dfrac{3x-1}{3}=\\dfrac{2-x}{2}$","$4(x-2)+3(x-4)=7x-20$ ","$x^2-2x+1=-4$"],"right":["$\\mathbb R$","$\\varnothing$","$S=\\{7\\}$","$S=\\left\\{\\dfrac {8}{9}\\right\\}$"],"top":100,"hint":"","explain":"<span class='basic_left'>$\\bullet\\,\\, 7-(2x+4)=-(x+4)\\\\ \\Leftrightarrow 7-2x-4=-x-4\\\\ \\Leftrightarrow x=7$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{7\\}$<br\/>$\\bullet\\,\\, \\dfrac{3x-1}{3}=\\dfrac{2-x}{2}\\\\ \\Leftrightarrow 2(3x-1)=3(2-x)\\\\ \\Leftrightarrow 6x-2=6-3x\\\\ \\Leftrightarrow 9x=8\\\\ \\Leftrightarrow x=\\dfrac {8}{9}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{\\dfrac {8}{9}\\right\\}$<br\/>$\\bullet \\,\\, 4(x-2)+3(x-4)=7x-20\\\\ \\Leftrightarrow 4x-8+3x-12=7x-20\\\\ \\Leftrightarrow 7x-20=7x-20\\,\\text{(lu\u00f4n \u0111\u00fang)}$<br\/> V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\mathbb R$<br\/>$\\bullet\\,\\, x^2-2x+1=-4\\\\ \\Leftrightarrow (x-1)^2=-4\\,\\text {(v\u00f4 nghi\u1ec7m)}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\varnothing$<br\/><\/span>"}]}],"id_ques":1014},{"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/bai22/lv2/img\/16.jpg' \/><\/center>T\u00ecm \u0111i\u1ec1u ki\u1ec7n c\u1ee7a $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh $(m-2)x-m+1=0$ l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n.","select":["A. $m=2$","B. $m\\ne 0$","C. $m\\ne 2$","D. $m=0$"],"hint":"","explain":" <span class='basic_left'>Ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 d\u1ea1ng $ax +b=0$, trong \u0111\u00f3: $a$ v\u00e0 $b$ l\u00e0 s\u1ed1 cho tr\u01b0\u1edbc v\u00e0 $a\\ne 0$<br\/>Do v\u1eady, \u0111\u1ec3 $(m-2)x-m+1=0$ l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n th\u00ec $m-2\\ne 0\\Leftrightarrow m\\ne 2$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":1015},{"time":24,"part":[{"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":[[["11"],["27"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"frac","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{2(3x+5)}{3}-\\dfrac{x}{2}=5-\\dfrac{3(x+1)}{4}.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>}","hint":"Quy \u0111\u1ed3ng v\u00e0 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n.","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$\\begin{align} & \\frac{2(3x+5)}{3}-\\frac{x}{2}=5-\\frac{3(x+1)}{4} \\\\ & \\Leftrightarrow 8(3x+5)-6x=60-9(x+1) \\\\ & \\Leftrightarrow 24x+40-6x=60-9x-9 \\\\ & \\Leftrightarrow 24x-6x+9x=60-9-40 \\\\ & \\Leftrightarrow 27x=11 \\\\ & \\Leftrightarrow x=\\dfrac{11}{27} \\\\ \\end{align}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{\\dfrac {11}{27}\\right\\}$<\/span>"}]}],"id_ques":1016},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. {$-4; \\dfrac{10}{9}$}","B. {$1; \\dfrac{-5}{3}$}","C. {$2; \\dfrac{5}{3}$}"],"ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $(5x-3)^2=(4x-7)^2.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${?;?}","hint":"Chuy\u1ec3n v\u1ebf. \u00c1p d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c $a^2-b^2$","explain":"<span class='basic_left'>Ta c\u00f3:<br\/>$(5x-3)^2=(4x-7)^2\\\\ \\Leftrightarrow (5x-3)^2-(4x-7)^2=0\\\\ \\Leftrightarrow [(5x-3)-(4x-7)][(5x-3)+(4x-7)]=0\\\\ \\Leftrightarrow (x+4)(9x-10)=0\\\\ \\Leftrightarrow \\left [\\begin{align}&x=-4\\\\ &x=\\dfrac{10}{9}\\\\ \\end{align}\\right.$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{-4;\\dfrac {10}{9}\\right\\}$<\/span>"}]}],"id_ques":1017},{"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/bai22/lv2/img\/8.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x^2-25=(x-5)(2x-11)$.<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{16\\}$","B. $S=\\{-16;5\\}$","C. $S=\\{5;16\\}$","D. $S=\\{-2;5\\}$"],"hint":"S\u1eed d\u1ee5ng h\u1eb1ng \u0111\u1eb3ng th\u1ee9c ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed.<br\/>\u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch","explain":" <span class='basic_left'>Ta c\u00f3: <br\/>$(x^2-25)=(x-5)(2x-11)\\\\ \\Leftrightarrow (x-5)(x+5)=(x-5)(2x-11)\\\\ \\Leftrightarrow (x-5)(x+5)-(x-5)(2x-11)=0\\\\ \\Leftrightarrow (x-5)[(x+5)-(2x-11)]=0\\\\ \\Leftrightarrow (x-5)(-x+16)=0\\\\ \\Leftrightarrow \\left[\\begin{align}&x=5\\\\&x=16\\\\ \\end{align}\\right.$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{5;16\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span>","column":2}]}],"id_ques":1018},{"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":[[["-9"],["2"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $x^2+7x-18=0.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_;_input_}","hint":"Ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed. \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch.","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$x^2+7x-18=0\\\\ \\Leftrightarrow x^2-2x+9x-18=0\\\\ \\Leftrightarrow x(x-2)+9(x-2)=0\\\\ \\Leftrightarrow (x-2)(x+9)=0\\\\ \\Leftrightarrow \\left [\\begin{align} &x=2\\\\ &x=-9\\\\ \\end{align}\\right.$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-9;2\\}$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-9$ v\u00e0 $2$<\/span><\/span>"}]}],"id_ques":1019},{"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/bai22/lv2/img\/10.jpg' \/><\/center>$S=\\{-1\\}$ l\u00e0 t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $x^3+x^2+3x+3=0$. <b>\u0110\u00fang<\/b> hay <b>Sai?<\/b>","select":["\u0110\u00fang","Sai"],"hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p nh\u00f3m v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung \u0111\u1ec3 ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed v\u00e0 \u0111\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh t\u00edch","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$x^3+x^2+3x+3=0\\\\ \\Leftrightarrow (x^3+x^2)+(3x+3)=0\\\\ \\Leftrightarrow x^2(x+1)+3(x+1)=0\\\\ \\Leftrightarrow (x+1)(x^2+3)=0\\\\ \\Leftrightarrow \\left[\\begin{aligned}&x=-1\\\\ &x^2=-3\\,\\,\\text{(v\u00f4 nghi\u1ec7m)}\\\\ \\end{aligned}\\right.$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-1\\}$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0111\u00fang<\/span><\/span>","column":2}]}],"id_ques":1020},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["-2","2"],["2","-2"],["1"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{x+1}{x-2}-\\dfrac{x-1}{x+2}=\\dfrac{2({{x}^{2}}+2)}{{{x}^{2}}-4}$ <br\/>a. X\u00e1c \u0111\u1ecbnh \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<br\/>b. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> <br\/>a. \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ v\u00e0 $x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<br\/>b. T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$<\/span>","hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 2$ v\u00e0 $x\\ne -2$<br\/>Ta c\u00f3:<br\/>$\\dfrac{x+1}{x-2}-\\dfrac{x-1}{x+2}=\\dfrac{2({{x}^{2}}+2)}{{{x}^{2}}-4}\\\\ \\Leftrightarrow \\dfrac{(x+1)(x+2)}{(x-2)(x+2)}-\\dfrac{(x-1)(x-2)}{(x+2)(x-2)}=\\dfrac{2({{x}^{2}}+2)}{(x-2)(x+2)}\\\\ \\Rightarrow (x+1)(x+2)-(x-1)(x-2)=2(x^2+2)\\\\ \\Leftrightarrow x^2+3x+2-(x^2-3x+2)-2x^2-4=0\\\\ \\Leftrightarrow -2x^2+6x-4=0\\\\ \\Leftrightarrow x^2-3x+2=0\\\\ \\Leftrightarrow x^2-x-2x+2=0\\\\ \\Leftrightarrow x(x-1)-2(x-1)=0\\\\ \\Leftrightarrow (x-1)(x-2)=0\\\\ \\Leftrightarrow\\left[\\begin{align}&x=1\\\\&x=2\\,\\,\\text{(lo\u1ea1i)}\\\\ \\end{align}\\right.$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{1\\}$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-2;2$ v\u00e0 $1$<\/span> <\/span>"}]}],"id_ques":1021},{"time":24,"part":[{"title":"Ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"select":["A. {$\\sqrt {5}-1;-\\sqrt {5}-1$}","B. {$\\sqrt {5}+1;-\\sqrt {5}-1$}","C. {$2\\sqrt {5}-1;-\\sqrt {5}-1$}"],"ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{-7{{x}^{2}}+6}{{{x}^{3}}+1}=\\dfrac{5}{{{x}^{2}}-x+1}-\\dfrac{3}{x+1}$<br\/><br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=?$<\/span>","hint":"","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne-1$ <br\/>Ta c\u00f3:<br\/>$\\dfrac{-7{{x}^{2}}+6}{{{x}^{3}}+1}=\\dfrac{5}{{{x}^{2}}-x+1}-\\dfrac{3}{x+1}\\\\ \\Leftrightarrow \\dfrac{-7{{x}^{2}}+6}{(x+1)(x^2-x+1)}=\\dfrac {5(x+1)}{(x+1)(x^2-x+1)}-\\dfrac {3(x^2-x+1)}{(x+1)(x^2-x+1)}\\\\ \\Rightarrow -7x^2+6=5x+5-3x^2+3x-3\\\\ \\Leftrightarrow 4x^2+8x-4=0\\\\ \\Leftrightarrow x^2+2x+1-5=0\\\\ \\Leftrightarrow (x+1)^2=5\\\\ \\Leftrightarrow \\left[\\begin{aligned} & x+1=\\sqrt {5}\\\\ &x+1=-\\sqrt{5}\\\\ \\end{aligned}\\right.\\Leftrightarrow \\left[\\begin{aligned} & x=\\sqrt {5}-1\\\\ &x =-\\sqrt{5}-1 \\\\ \\end{aligned}\\right.$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\left\\{\\sqrt {5}-1;-\\sqrt {5}-1\\right\\}$<\/span>"}]}],"id_ques":1022},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\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/bai22/lv2/img\/9.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh tr\u00ecnh $\\dfrac{x-1}{x+3}-\\dfrac{x}{x-3}=\\dfrac{7x-3}{9-{{x}^{2}}}$ c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\mathbb R .$","select":["\u0110\u00fang ","Sai "],"hint":"Quy \u0111\u1ed3ng r\u1ed3i gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","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\/>$\\dfrac{x-1}{x+3}-\\dfrac{x}{x-3}=\\dfrac{7x-3}{9-{{x}^{2}}}\\\\ \\Leftrightarrow \\dfrac {(x-1)(x-3)}{(x-3)(x+3)}-\\dfrac {x(x+3)}{(x-3)(x+3)}=\\dfrac{3-7x}{(x-3)(x+3)}\\\\ \\Rightarrow (x-1)(x-3)-x(x+3)=3-7x\\\\ \\Leftrightarrow x^2-4x+3-x^2-3x-3+7x=0\\\\ \\Leftrightarrow 0=0\\,\\,\\text{(lu\u00f4n \u0111\u00fang)}$ <br\/>K\u1ebft h\u1ee3p v\u1edbi \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh, ta c\u00f3: T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 t\u1eadp c\u00e1c s\u1ed1 th\u1ef1c kh\u00e1c $3$ v\u00e0 kh\u00e1c $-3$.<br\/>Ta th\u01b0\u1eddng k\u00ed hi\u1ec7u l\u00e0 $\\mathbb R $\\$ \\{3;-3\\}$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai<\/span><br\/><b> Ghi nh\u1edb: <\/b> V\u1edbi ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu, t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n lo\u1ea1i b\u1ecf nh\u1eefng gi\u00e1 tr\u1ecb nghi\u1ec7m kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh.<br\/>Do v\u1eady, khi gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh th\u1ea5y c\u00f3 v\u00f4 s\u1ed1 nghi\u1ec7m c\u1ea7n lo\u1ea1i b\u1ecf nh\u1eefng nghi\u1ec7m kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh.<br\/><span class='basic_green'>L\u01b0u \u00fd: <\/span>K\u00ed hi\u1ec7u $\\mathbb R $\\$\\{a;b\\}$ \u0111\u01b0\u1ee3c hi\u1ec3u l\u00e0 t\u1eadp h\u1ee3p c\u00e1c s\u1ed1 th\u1ef1c kh\u00e1c $a$ v\u00e0 $b$<\/span>","column":2}]}],"id_ques":1023},{"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":[[["0"],["-5"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $(x^2+5x+1)(x^2+5x+9)=9.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_;_input_}","hint":"\u0110\u1eb7t $x^2+5x=t$","explain":"<span class='basic_left'>\u0110\u1eb7t $x^2+5x=t$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u tr\u1edf th\u00e0nh: <br\/>$(t+1)(t+9)=9\\\\ \\Leftrightarrow t^2+10t+9=9\\\\ \\Leftrightarrow t^2+10t=0\\\\ \\Leftrightarrow t(t+10)=0 \\\\ \\Leftrightarrow\\left[ \\begin {aligned}&t=0\\\\ &t=-10\\\\ \\end{aligned} \\right.$<br\/>V\u1edbi $t=0$, ta c\u00f3:<br\/>$x^2+5x=0\\\\ \\Leftrightarrow x(x+5)=0\\\\ \\Leftrightarrow\\left[\\begin{aligned}&x=0\\\\ &x=-5\\\\ \\end{aligned}\\right.$<br\/>V\u1edbi $t=-10$, ta c\u00f3:<br\/>$x^2+5x+10=0\\\\ \\Leftrightarrow x^2+2.x.\\dfrac {5}{2}+\\dfrac{25}{4}+\\dfrac{15}{4}=0\\\\ \\Leftrightarrow \\left(x+\\dfrac{5}{2}\\right)^2=-\\dfrac{15}{4}\\,\\,\\text{(v\u00f4 nghi\u1ec7m)}$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{0;-5\\}$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-5;0$<\/span><\/span>"}]}],"id_ques":1024},{"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":[[["0"],["4"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{5}{{{x}^{2}}-4x+5}-({{x}^{2}}-4x+1)=0.$ <br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_;_input_}","hint":"\u0110\u1eb7t $x^2-4x+5=t\\Rightarrow x^2-4x+1=t-4$","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x^2-4x+5\\ne 0\\Leftrightarrow x^2-4x+4+1\\ne 0\\Leftrightarrow (x-2)^2+1\\ne 0\\,\\forall x$<br\/>\u0110\u1eb7t $x^2-4x+5=t\\Rightarrow x^2-4x+1=t-4$ ($t > 0$)<br\/>Ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u tr\u1edf th\u00e0nh: <br\/>$\\dfrac{5}{t}-(t-4)=0\\\\ \\Leftrightarrow \\dfrac{5}{t}-\\dfrac{t(t-4)}{t}=0\\\\ \\Rightarrow 5-(t^2-4t)=0\\\\ \\Leftrightarrow t^2-4t-5=0\\\\\\Leftrightarrow t^2-5t+t-5=0\\\\ \\Leftrightarrow t(t-5)+(t-5)=0\\\\ \\Leftrightarrow (t+1)(t-5)=0\\\\ \\Leftrightarrow\\left[\\begin{align}& t=-1\\,\\,\\,\\text{(lo\u1ea1i)}\\\\&t=5\\,\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ \\end{align}\\right.$<br\/>V\u1edbi $t=5$, ta c\u00f3:<br\/>$x^2-4x+5=5\\\\ \\Leftrightarrow x^2-4x=0\\\\ \\Leftrightarrow x(x-4)=0\\\\ \\Leftrightarrow \\left[\\begin{align} &x=0\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\&x=4\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ \\end{align}\\right.$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{0;4\\}$<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0;4$<\/span><\/span>"}]}],"id_ques":1025},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":5,"width":50,"type_input":"","ques":"Ph\u01b0\u01a1ng tr\u00ecnh $3y^3-7y^2-7y+3=0$ c\u00f3 bao nhi\u00eau nghi\u1ec7m?","select":["A. $1$ nghi\u1ec7m","B. $2$ nghi\u1ec7m","C. $3$ nghi\u1ec7m","D. v\u00f4 nghi\u1ec7m"],"hint":"S\u1eed d\u1ee5ng ph\u01b0\u01a1ng ph\u00e1p nh\u00f3m v\u00e0 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c ph\u00e2n t\u00edch v\u1ebf tr\u00e1i th\u00e0nh nh\u00e2n t\u1eed.","explain":"<span class='basic_left'>Ta c\u00f3: <br\/>$3y^3-7y^2-7y+3=0\\\\ \\Leftrightarrow (3y^3+3)-(7y^2+7y)=0\\\\ \\Leftrightarrow 3(y^3+1)-7y(y+1)=0\\\\ \\Leftrightarrow 3(y+1)(y^2-y+1)-7y(y+1)=0\\\\ \\Leftrightarrow (y+1)(3y^2-3y+3-7y)=0\\\\ \\Leftrightarrow (y+1)(3y^2-10y+3)=0\\\\ \\Leftrightarrow (y+1)(3y^2-9y-y+3)=0\\\\ \\Leftrightarrow (y+1)(y-3)(3y-1)=0\\\\ \\Leftrightarrow \\left[\\begin{aligned} &y=-1\\\\ &y=3\\\\ &y=\\dfrac{1}{3}\\\\ \\end{aligned}\\right.$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 $3$ nghi\u1ec7m $y\\in\\left\\{-1;3;\\dfrac{1}{3}\\right\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C.<\/span><\/span>","column":2}]}],"id_ques":1026},{"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/bai22/lv2/img\/16.jpg' \/><\/center>H\u1ecdc k\u1ef3 $1$, s\u1ed1 h\u1ecdc sinh gi\u1ecfi l\u1edbp $8A$ b\u1eb1ng $\\dfrac {1}{6}$ s\u1ed1 h\u1ecdc sinh c\u1ea3 l\u1edbp. Sang h\u1ecdc k\u00ec $2$, c\u00f3 th\u00eam $4$ b\u1ea1n t\u1eeb l\u1edbp $8B$ chuy\u1ec3n sang v\u00e0 c\u00f3 $2$ b\u1ea1n ph\u1ea5n \u0111\u1ea5u tr\u1edf th\u00e0nh h\u1ecdc sinh gi\u1ecfi n\u1eefa n\u00ean s\u1ed1 h\u1ecdc sinh gi\u1ecfi b\u1eb1ng $\\dfrac {1}{5}$ s\u1ed1 h\u1ecdc sinh c\u1ea3 l\u1edbp. T\u00ecm s\u1ed1 h\u1ecdc sinh l\u1edbp $8A$ \u1edf h\u1ecdc k\u00ec $2$.","select":["A. $18$ h\u1ecdc sinh","B. $24$ h\u1ecdc sinh","C. $36$ h\u1ecdc sinh","D. $40$ h\u1ecdc sinh"],"hint":"G\u1ecdi s\u1ed1 h\u1ecdc sinh c\u1ea3 l\u1edbp \u1edf h\u1ecdc k\u00ec $1$ l\u00e0 \u1ea9n s\u1ed1, bi\u1ec3u di\u1ec5n s\u1ed1 h\u1ecdc sinh gi\u1ecfi theo \u1ea9n.","explain":" <span class='basic_left'>G\u1ecdi s\u1ed1 h\u1ecdc sinh c\u1ea3 l\u1edbp l\u00e0 $x$ (h\u1ecdc sinh, $x\\in\\mathbb N^*$)<br\/>H\u1ecdc k\u00ec $1$, s\u1ed1 h\u1ecdc sinh gi\u1ecfi b\u1eb1ng $\\dfrac {1}{6}$ h\u1ecdc sinh c\u1ea3 l\u1edbp n\u00ean s\u1ed1 h\u1ecdc sinh gi\u1ecfi l\u00e0 $\\dfrac {x}{6}$ (h\u1ecdc sinh)<br\/>H\u1ecdc k\u00ec $2$, l\u1edbp $8A$ c\u00f3 $x+4$ (h\u1ecdc sinh).<br\/>V\u00ec c\u00f3 th\u00eam $2$ b\u1ea1n h\u1ecdc sinh gi\u1ecfi n\u00ean s\u1ed1 h\u1ecdc sinh gi\u1ecfi h\u1ecdc k\u00ec $2$ l\u00e0 $\\dfrac{x}{6}+2$ (h\u1ecdc sinh)<br\/>V\u00ec s\u1ed1 h\u1ecdc sinh gi\u1ecfi h\u1ecdc k\u00ec $2$ b\u1eb1ng $\\dfrac {1}{5}$ s\u1ed1 h\u1ecdc sinh c\u1ea3 l\u1edbp n\u00ean s\u1ed1 h\u1ecdc sinh gi\u1ecfi h\u1ecdc k\u00ec $2$ l\u00e0 $\\dfrac {x+4}{5}$ (h\u1ecdc sinh)<br\/> Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\dfrac{x}{6}+2=\\dfrac {x+4}{5}\\\\ \\Leftrightarrow 5x+60=6(x+4)\\\\ \\Leftrightarrow 5x+60=6x+24\\\\ \\Leftrightarrow x=36 \\,\\,\\text{(th\u1ecfa m\u00e3n)}$<br\/>V\u1eady h\u1ecdc k\u00ec $1$, l\u1edbp $8A$ c\u00f3 $36$ b\u1ea1n h\u1ecdc sinh.<br\/>Do v\u1eady h\u1ecdc k\u00ec $2$, l\u1edbp $8A$ c\u00f3 $40$ h\u1ecdc sinh.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":1027},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["350"],["8"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai22/lv2/img\/11.jpg' \/><\/center><span class='basic_left'>M\u1ed9t \u00f4 t\u00f4 d\u1ef1 \u0111\u1ecbnh \u0111i t\u1eeb $A$ \u0111\u1ebfn $B$. N\u1ebfu \u00f4 t\u00f4 \u0111i v\u1edbi v\u1eadn t\u1ed1c $35 km\/h$ th\u00ec \u00f4 t\u00f4 \u0111\u1ebfn mu\u1ed9n $2$ gi\u1edd. N\u1ebfu \u00f4 t\u00f4 \u0111i v\u1edbi v\u1eadn t\u1ed1c $50 km\/h$ th\u00ec \u00f4 t\u00f4 \u0111\u1ebfn s\u1edbm h\u01a1n $1$ gi\u1edd. T\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng $AB$ v\u00e0 th\u1eddi gian d\u1ef1 \u0111\u1ecbnh.<br\/><b>\u0110\u00e1p s\u1ed1: <\/b> <br\/>Qu\u00e3ng \u0111\u01b0\u1eddng $AB$ l\u00e0 $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}(km)$;<br\/>Th\u1eddi gian d\u1ef1 \u0111\u1ecbnh l\u00e0 $\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$(gi\u1edd)<\/span>","hint":"L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh d\u1ef1a tr\u00ean m\u1ed1i li\u00ean h\u1ec7 v\u1ec1 th\u1eddi gian","explain":"<span class='basic_left'>G\u1ecdi \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng $AB$ l\u00e0 $x$ $(km, x>0)$<br\/>N\u1ebfu \u00f4 t\u00f4 \u0111i v\u1edbi v\u1eadn t\u1ed1c $35 km\/h$ th\u00ec h\u1ebft $\\dfrac {x}{35}$ (gi\u1edd)<br\/>V\u00ec \u00f4 t\u00f4 \u0111i v\u1edbi v\u1eadn t\u1ed1c $35 km\/h$ th\u00ec \u00f4 t\u00f4 \u0111\u1ebfn mu\u1ed9n $2$ gi\u1edd n\u00ean th\u1eddi gian d\u1ef1 \u0111\u1ecbnh l\u00e0 $\\dfrac {x}{35}-2$ (gi\u1edd)<br\/>N\u1ebfu \u00f4 t\u00f4 \u0111i v\u1edbi v\u1eadn t\u1ed1c $50km\/h$ th\u00ec h\u1ebft $\\dfrac {x}{50}$ (gi\u1edd)<br\/>V\u00ec \u00f4 t\u00f4 \u0111i v\u1edbi v\u1eadn t\u1ed1c $50 km\/h$ th\u00ec \u00f4 t\u00f4 \u0111\u1ebfn s\u1edbm h\u01a1n $1$ gi\u1edd n\u00ean th\u1eddi gian d\u1ef1 \u0111\u1ecbnh l\u00e0 $\\dfrac {x}{50}+1$ (gi\u1edd)<br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\dfrac {x}{35}-2=\\dfrac {x}{50}+1\\\\ \\Leftrightarrow 10x=7x+1050\\\\ \\Leftrightarrow x=350 \\,\\,\\,\\text{(th\u1ecfa m\u00e3n)} $<br\/>V\u1eady qu\u00e3ng \u0111\u01b0\u1eddng $AB$ d\u00e0i $350 km$. <br\/>Khi \u0111\u00f3, th\u1eddi gian d\u1ef1 \u0111\u1ecbnh m\u00e0 \u00f4 t\u00f4 \u0111i l\u00e0 $\\dfrac {350}{35}-2=8$ gi\u1edd<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u1ea7n l\u01b0\u1ee3t l\u00e0 $350$ v\u00e0 $8$ <\/span><\/span>"}]}],"id_ques":1028},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["300"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"M\u1ed9t x\u00ed nghi\u1ec7p k\u00fd h\u1ee3p \u0111\u1ed3ng d\u1ec7t m\u1ed9t s\u1ed1 t\u1ea5m th\u1ea3m len trong $20$ ng\u00e0y. Do c\u1ea3i ti\u1ebfn k\u0129 thu\u1eadt, n\u0103ng su\u1ea5t d\u1ec7t c\u1ee7a x\u00ed nghi\u1ec7p \u0111\u00e3 t\u0103ng $20\\%$. B\u1edfi v\u1eady, ch\u1ec9 trong $18$ ng\u00e0y, kh\u00f4ng nh\u1eefng x\u00ed nghi\u1ec7p \u0111\u00e3 ho\u00e0n th\u00e0nh s\u1ed1 th\u1ea3m c\u1ea7n d\u1ec7t m\u00e0 c\u00f2n d\u1ec7t th\u00eam \u0111\u01b0\u1ee3c $24$ t\u1ea5m n\u1eefa. T\u00ednh s\u1ed1 t\u1ea5m th\u1ea3m len m\u00e0 x\u00ed nghi\u1ec7p ph\u1ea3i d\u1ec7t theo h\u1ee3p \u0111\u1ed3ng.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> _input_ (t\u1ea5m th\u1ea3m)","hint":"B\u00e0i to\u00e1n n\u0103ng su\u1ea5t: T\u1ed5ng s\u1ed1 t\u1ea5m th\u1ea3m = n\u0103ng su\u1ea5t * s\u1ed1 ng\u00e0y","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>L\u1eadp b\u1ea3ng:<br\/><table><tr><th><\/th><th>T\u1ed5ng s\u1ed1 th\u1ea3m<\/th><th>N\u0103ng su\u1ea5t<\/th><th>S\u1ed1 ng\u00e0y<\/th><\/tr><tr><td>Theo h\u1ee3p \u0111\u1ed3ng<\/td><td>$x$<\/td><td>$\\dfrac{x}{20}$<\/td><td>$20$<\/td><\/tr><tr><td>Th\u1ef1c t\u1ebf<\/td><td>$x+24$<\/td><td> $\\dfrac{x}{20}+\\dfrac{x}{20}.20\\%=\\dfrac {3x}{50}$ <\/td><td>$18$<\/td><\/tr><\/table><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi s\u1ed1 th\u1ea3m len m\u00e0 x\u00ed nghi\u1ec7p ph\u1ea3i d\u1ec7t theo h\u1ee3p \u0111\u1ed3ng l\u00e0 $x$ (t\u1ea5m v\u1ea3i, $x \\in \\mathbb {N^*}$)<br\/>Theo d\u1ef1 \u0111\u1ecbnh m\u1ed7i ng\u00e0y x\u00ed nghi\u1ec7p ph\u1ea3i d\u1ec7t $\\dfrac{x}{20}$ t\u1ea5m v\u1ea3i<br\/>Do c\u1ea3i ti\u1ebfn k\u0129 thu\u1eadt, n\u0103ng su\u1ea5t d\u1ec7t c\u1ee7a x\u00ed nghi\u1ec7p \u0111\u00e3 t\u0103ng $20\\%$ n\u00ean s\u1ed1 t\u1ea5m th\u1ea3m m\u00e0 th\u1ef1c t\u1ebf x\u00ed nghi\u1ec7p \u0111\u00e3 d\u1ec7t l\u00e0 <br\/>$\\dfrac{x}{20}+\\dfrac{x}{20}.20\\%=\\dfrac {3x}{50}$ (t\u1ea5m th\u1ea3m)<br\/>V\u00ec x\u00ed nghi\u1ec7p l\u00e0m trong $18$ ng\u00e0y n\u00ean s\u1ed1 t\u1ea5m v\u1ea3i th\u1ef1c t\u1ebf \u0111\u00e3 l\u00e0m \u0111\u01b0\u1ee3c l\u00e0 $18.\\dfrac {3x}{50}=\\dfrac {27x}{25}$ (t\u1ea5m th\u1ea3m)<br\/>V\u00ec x\u00ed nghi\u1ec7p d\u1ec7t th\u00eam \u0111\u01b0\u1ee3c $24$ t\u1ea5m n\u1eefa n\u00ean s\u1ed1 t\u1ea5m v\u1ea3i x\u00ed nghi\u1ec7p th\u1ef1c t\u1ebf l\u00e0m \u0111\u01b0\u1ee3c l\u00e0 $x+24$ (t\u1ea5m th\u1ea3m)<br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\dfrac {27x}{25}=x+24\\\\ \\Leftrightarrow 27x=25x+600\\\\ \\Leftrightarrow 2x=600\\\\ \\Leftrightarrow x=300\\,\\text{(th\u1ecfa m\u00e3n)}$<br\/>V\u1eady s\u1ed1 t\u1ea5m th\u1ea3m len m\u00e0 x\u00ed nghi\u1ec7p ph\u1ea3i d\u1ec7t theo h\u1ee3p \u0111\u1ed3ng l\u00e0 $300$ t\u1ea5m<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $300$<\/span><\/span>"}]}],"id_ques":1029},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["150"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"M\u1ed9t ng\u01b0\u1eddi \u0111i xe m\u00e1y t\u1eeb $A$ \u0111\u1ebfn $B$ d\u1ef1 \u0111\u1ecbnh m\u1ea5t $3$ gi\u1edd $20$ ph\u00fat. N\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 t\u0103ng v\u1eadn t\u1ed1c th\u00eam $5km\/h$ th\u00ec ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ebfn $B$ s\u1edbm h\u01a1n d\u1ef1 \u0111\u1ecbnh $20$ ph\u00fat. T\u00ednh qu\u00e3ng \u0111\u01b0\u1eddng $AB$<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> _input_ ($km$)","hint":"Qu\u00e3ng \u0111\u01b0\u1eddng $AB$ l\u00e0 \u1ea9n s\u1ed1, l\u1eadp ph\u01b0\u01a1ng tr\u00ecnh d\u1ef1a tr\u00ean \u0111\u1ed9 d\u00e0i qu\u00e3ng \u0111\u01b0\u1eddng $AB$ kh\u00f4ng \u0111\u1ed5i.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/>L\u1eadp b\u1ea3ng:<br\/><table><tr><th><\/th><th>Qu\u00e3ng \u0111\u01b0\u1eddng $(km)$<\/th><th>V\u1eadn t\u1ed1c ($km\/h$)<\/th><th>Th\u1eddi gian (gi\u1edd)<\/th><\/tr><tr><td>Theo d\u1ef1 \u0111\u1ecbnh<\/td><td>$x$<\/td><td>$x:\\dfrac {10}{3}=\\dfrac {3x}{10}$<\/td><td>$3$ gi\u1edd $20$ ph\u00fat $=\\dfrac {10}{3}$ gi\u1edd<\/td><\/tr><tr><td>Th\u1ef1c t\u1ebf<\/td><td>$x$<\/td><td> $\\dfrac {3x}{10}+5$ <\/td><td>$3$ <\/td><\/tr><\/table><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi qu\u00e3ng \u0111\u01b0\u1eddng $AB$ l\u00e0 $x$ ($km$, $x>0$)<br\/>V\u00ec d\u1ef1 \u0111\u1ecbnh ng\u01b0\u1eddi \u0111\u00f3 \u0111i \u0111\u1ebfn $B$ m\u1ea5t $3$ gi\u1edd $20$ ph\u00fat $=\\dfrac {10}{3}$ gi\u1edd n\u00ean v\u1eadn t\u1ed1c theo d\u1ef1 \u0111\u1ecbnh l\u00e0 $x:\\dfrac {10}{3}=\\dfrac {3x}{10}$ ($km\/h$)<br\/>Khi v\u1eadn t\u1ed1c t\u0103ng $5\\, km\/h$ th\u00ec v\u1eadn t\u1ed1c l\u00e0 $\\dfrac {3x}{10}+5$ ($km\/h$)<br\/>V\u00ec ng\u01b0\u1eddi \u0111\u00f3 \u0111\u1ebfn $B$ s\u1edbm h\u01a1n d\u1ef1 \u0111\u1ecbnh $20$ ph\u00fat l\u00e0 $3$ gi\u1edd n\u00ean qu\u00e3ng \u0111\u01b0\u1eddng $AB$ l\u00e0 $\\left(\\dfrac {3x}{10}+5\\right).3$ ($km$)<br\/>Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\left(\\dfrac {3x}{10}+5\\right).3=x\\\\ \\Leftrightarrow \\dfrac {9x}{10}+15=x\\\\ \\Leftrightarrow 9x+150=10x\\\\ \\Leftrightarrow x=150\\,\\text{(th\u1ecfa m\u00e3n)}$<br\/>V\u1eady qu\u00e3ng \u0111\u01b0\u1eddng $AB$ d\u00e0i $150 km$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $150$<\/span><\/span>"}]}],"id_ques":1030}],"lesson":{"save":0,"level":2}}