{"segment":[{"time":24,"part":[{"time":3,"title":"N\u1ed1i m\u1ed7i ph\u01b0\u01a1ng tr\u00ecnh c\u1ed9t tr\u00e1i v\u1edbi ph\u01b0\u01a1ng tr\u00ecnh b\u1eadc nh\u1ea5t m\u1ed9t \u1ea9n t\u01b0\u01a1ng \u0111\u01b0\u01a1ng v\u1edbi n\u00f3 \u1edf c\u1ed9t ph\u1ea3i ","title_trans":"","audio":"","temp":"matching","correct":[["2","3","1"]],"list":[{"point":10,"image":"","left":["$5(x+3)^2-5(x-4)(x+8)=0$ ","$(2x-1)(4x^2+2x+1)\\\\-4x(2x^2-3)=23$","$x(x^2+x+1)-(x^2-1)x=x^2+2$"],"right":["$x-1=0$","$2x+41=0$","$x-2=0$"],"top":105,"hint":"Bi\u1ebfn \u0111\u1ed5i t\u01b0\u01a1ng \u0111\u01b0\u01a1ng m\u1ed7i ph\u01b0\u01a1ng tr\u00ecnh.","explain":"<span class='basic_left'>$\\bullet\\,\\, 5(x+3)^2-5(x-4)(x+8)=0\\\\ \\Leftrightarrow 5(x^2+6x+9)-5(x^2+4x-32)=0\\\\ \\Leftrightarrow 5x^2+30x+45-5x^2-20x+160=0\\\\ \\Leftrightarrow 10x+205=0\\\\ \\Leftrightarrow 2x+41=0$ <br\/> $\\bullet\\,\\, (2x-1)(4x^2+2x+1)-4x(2x^2-3)=23\\\\ \\Leftrightarrow (2x)^3-1-8x^3+12x-23=0\\\\ \\Leftrightarrow 12x-24=0\\\\ \\Leftrightarrow x-2=0$ <br\/> $\\bullet\\,\\, x(x^2+x+1)-(x^2-1)x=x^2+2\\\\ \\Leftrightarrow x^3+x^2+x-(x^2-1)x-x^2-2=0\\\\ \\Leftrightarrow x^3+x^2+x-x^3+x-x^2-2=0\\\\ \\Leftrightarrow 2x-2=0\\\\ \\Leftrightarrow x-1=0$<br\/><\/span>"}]}],"id_ques":1031},{"time":24,"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","4"]],"list":[{"point":10,"img":"","ques":"Trong c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh sau ph\u01b0\u01a1ng tr\u00ecnh n\u00e0o v\u00f4 nghi\u1ec7m?","hint":"","column":1,"number_true":2,"select":["A. $2x+1=3x+2$","B. $\\dfrac{2x}{x-1}-2=\\dfrac{3}{1-x}$","C. $\\dfrac{x+1}{2}-\\dfrac{2(x+1)}{3}=0$","D. $ x^2+4x+18=0$","E. $x^2+2x=8$"],"explain":"<span class='basic_left'>A. $ 2x+1=3x+2\\\\ \\Leftrightarrow x=-1$<br\/>B. $\\dfrac{2x}{x-1}-2=\\dfrac{3}{1-x}\\,\\,\\text{(\u0110i\u1ec1u ki\u1ec7n:} x\\ne 1)\\\\ \\Leftrightarrow \\dfrac {2x-2(x-1)}{x-1}=\\dfrac {-3}{x-1}\\\\ \\Leftrightarrow \\dfrac {2}{x-1}=\\dfrac {-3}{x-1}\\\\ \\Leftrightarrow \\dfrac {5}{x-1}=0\\,\\, \\text{(v\u00f4 nghi\u1ec7m)}$<br\/>C. $\\dfrac{x+1}{2}-\\dfrac{2(x+1)}{3}=0\\\\ \\Leftrightarrow 3(x+1)-4(x+1)=0\\\\ \\Leftrightarrow -x-1=0\\\\ \\Leftrightarrow x=-1$<br\/>D. $ x^2+4x+18=0\\\\ \\Leftrightarrow x^2+4x+4+14=0\\\\ \\Leftrightarrow (x+2)^2=-14\\,\\text{(v\u00f4 nghi\u1ec7m)}$<br\/>E. $x^2+2x=8\\\\ \\Leftrightarrow x^2+2x+1=9\\\\ \\Leftrightarrow (x+1)^2=9\\\\ \\Leftrightarrow \\left[\\begin{aligned}&x+1=3\\\\ &x+1=-3\\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left[\\begin{aligned}&x=2\\\\ &x=-4\\\\ \\end{aligned}\\right.$<br\/><span class='basic_pink'>V\u1eady c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh v\u00f4 nghi\u1ec7m l\u00e0 B v\u00e0 D<\/span> <\/span>"}]}],"id_ques":1032},{"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"],["-2"],["1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai22/lv3/img\/9.jpg' \/><\/center><span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x+2}{x-m}=\\dfrac{x+1}{x-1}$ ($m$ l\u00e0 tham s\u1ed1). T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t.<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> $m\\ne\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$; $m\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ v\u00e0 $m\\ne\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span>","hint":"Quy \u0111\u1ed3ng ph\u01b0\u01a1ng tr\u00ecnh nh\u01b0 b\u00e0i to\u00e1n gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu, r\u1ed3i bi\u1ec7n lu\u1eadn s\u1ed1 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh theo $m$.","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne m$ v\u00e0 $x\\ne 1$<br\/>Ta c\u00f3: <br\/>$\\dfrac{x+2}{x-m}=\\dfrac{x+1}{x-1}\\,\\,(*)\\\\ \\Leftrightarrow \\dfrac {(x+2)(x-1)}{(x-m)(x-1)}=\\dfrac {(x+1)(x-m)}{(x-m)(x-1)}\\\\ \\Rightarrow (x+2)(x-1)=(x-m)(x+1)\\\\ \\Leftrightarrow x^2+x-2-(x^2-mx+x-m)=0\\\\\\Leftrightarrow x^2+x-2-x^2+mx-x+m=0\\\\ \\Leftrightarrow mx+m-2=0\\,\\,(1)$ <br\/>\u0110\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh (*) c\u00f3 nghi\u1ec7m duy nh\u1ea5t th\u00ec ph\u01b0\u01a1ng tr\u00ecnh (1) c\u00f3 m\u1ed9t nghi\u1ec7m duy nh\u1ea5t kh\u00e1c $m$ v\u00e0 kh\u00e1c $1$<br\/>Khi \u0111\u00f3, ta c\u00f3: <br\/>$\\left\\{\\begin{aligned} & m\\ne 0 \\\\ & \\dfrac {-m+2}{m}\\ne 1 \\\\ & \\dfrac {-m+2}{m}\\ne m \\\\ \\end{aligned} \\right. \\\\ \\Leftrightarrow \\left \\{ \\begin{aligned} & m\\ne 0 \\\\ & -m+2\\ne m \\\\ & -m+2\\ne m^2 \\\\ \\end{aligned}\\right. \\\\ \\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & 2m\\ne 2 \\\\ & m^2+m-2\\ne 0\\\\ \\end{aligned}\\right. \\\\ \\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & m\\ne 1 \\\\ & (m-1)(m+2)\\ne 0\\\\ \\end{aligned}\\right.\\\\ \\Leftrightarrow \\left\\{ \\begin{aligned} & m\\ne 0 \\\\ & m\\ne 1 \\\\ & m\\ne -2\\\\ \\end{aligned}\\right.$<br\/>V\u1eady \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 nghi\u1ec7m duy nh\u1ea5t th\u00ec $m\\ne 0$; $m\\ne 1$v\u00e0 $m\\ne -2$<br\/><span class='basic_pink'>V\u1eady c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0;1$ v\u00e0 $-2$<\/span><\/span>"}]}],"id_ques":1033},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai22/lv3/img\/8.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac {x+1}{65}+\\dfrac{x+3}{63}=\\dfrac{x+5}{61}+\\dfrac{x+7}{59}.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{66\\}$","B. $S=\\{-66\\}$","C. $S=\\{-1\\}$","D. $S=\\{1\\}$"],"hint":"V\u00ec \u1edf c\u00e1c ph\u00e2n th\u1ee9c, khi c\u1ed9ng t\u1eed th\u1ee9c v\u1edbi m\u1eabu th\u1ee9c v\u1edbi nhau ta thu \u0111\u01b0\u1ee3c c\u00e1c bi\u1ec3u th\u1ee9c gi\u1ed1ng nhau n\u00ean c\u1ed9ng m\u1ed7i ph\u00e2n th\u1ee9c v\u1edbi $1$ \u0111\u01a1n v\u1ecb \u0111\u1ec3 thu \u0111\u01b0\u1ee3c nh\u00e2n t\u1eed chung.","explain":" <span class='basic_left'>Ta c\u00f3:<br\/>$\\dfrac {x+1}{65}+\\dfrac{x+3}{63}=\\dfrac{x+5}{61}+\\dfrac{x+7}{59}\\\\ \\Leftrightarrow \\dfrac {x+1}{65}+1+\\dfrac{x+3}{63}+1=\\dfrac{x+5}{61}+1+\\dfrac{x+7}{59}+1\\\\ \\Leftrightarrow \\dfrac {x+1+65}{65}+\\dfrac{x+3+63}{63}=\\dfrac{x+5+61}{61}+\\dfrac{x+7+59}{59}\\\\ \\Leftrightarrow \\dfrac {x+66}{65}+\\dfrac{x+66}{63}=\\dfrac{x+66}{61}+\\dfrac{x+66}{59}\\\\ \\Leftrightarrow (x+66)\\left(\\dfrac {1}{65}+\\dfrac{1}{63}-\\dfrac{1}{61}-\\dfrac{1}{59}\\right)=0\\,\\,(*)$ <br\/>V\u00ec $\\dfrac {1}{65}+\\dfrac{1}{63}-\\dfrac{1}{61}-\\dfrac{1}{59}\\ne 0$. <br\/>$(*)\\Leftrightarrow x+66=0\\Leftrightarrow x=-66$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-66\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B <\/span><br\/><span class='basic_green'>Ghi nh\u1edb: <\/span> Ph\u01b0\u01a1ng tr\u00ecnh t\u1ed5ng qu\u00e1t d\u1ea1ng: $\\dfrac{f_1(x)}{g_1(x)}+\\dfrac{f_2(x)}{g_2(x)}+\\dfrac{f_3(x)}{g_3(x)}+...+\\dfrac{f_n(x)}{g_n(x)}=0$ <br\/>+) N\u1ebfu $f_1(x)+g_1(x)\\equiv f_2(x)+g_2(x)\\equiv f_3(x)+g_3\\equiv ...\\equiv f_n(x)+g_n(x)$ th\u00ec ta th\u1ef1c hi\u1ec7n gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eb1ng c\u00e1ch c\u1ed9ng m\u1ed7i ph\u00e2n th\u1ee9c v\u1edbi $1$ r\u1ed3i quy \u0111\u1ed3ng v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung.<br\/>+) N\u1ebfu $f_1(x)-g_1(x)\\equiv f_2(x)-g_2(x)\\equiv f_3(x)-g_3(x)\\equiv ...\\equiv f_n(x)-g_n(x)$ th\u00ec ta th\u1ef1c hi\u1ec7n gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eb1ng c\u00e1ch tr\u1eeb m\u1ed7i ph\u00e2n th\u1ee9c cho $1$ r\u1ed3i quy \u0111\u1ed3ng v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung.<\/span>","column":2}]}],"id_ques":1034},{"time":24,"part":[{"title":"Kh\u1eb3ng \u0111\u1ecbnh sau \u0111\u00fang hay sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai22/lv3/img\/16.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x}{2x+1}+\\dfrac{x-1}{2x+2}=\\dfrac{x-2}{2x+3}+\\dfrac{x-3}{2x+4}.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{-1\\}$","select":["\u0110\u00fang","Sai"],"hint":"V\u00ec \u1edf c\u00e1c ph\u00e2n th\u1ee9c, khi tr\u1eeb t\u1eed th\u1ee9c cho m\u1eabu th\u1ee9c ta thu \u0111\u01b0\u1ee3c c\u00e1c bi\u1ec3u th\u1ee9c gi\u1ed1ng nhau n\u00ean tr\u1eeb m\u1ed7i ph\u00e2n th\u1ee9c cho $1$ \u0111\u01a1n v\u1ecb \u0111\u1ec3 thu \u0111\u01b0\u1ee3c nh\u00e2n t\u1eed chung.","explain":"<span class='basic_left'> \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $\\left\\{\\begin{aligned}&x \\ne \\dfrac {-1}{2}\\\\ & x\\ne -1\\\\ &x\\ne \\dfrac {-3}{2}\\\\ & x\\ne -2\\\\ \\end{aligned}\\right.$<br\/>Ta c\u00f3:<br\/>$\\begin{align}& \\dfrac{x}{2x+1}+\\dfrac{x+1}{2x+2}=\\dfrac{x+2}{2x+3}+\\dfrac{x+3}{2x+4} \\\\ & \\Leftrightarrow \\dfrac{x}{2x+1}-1+\\dfrac{x-1}{2x+2}-1=\\dfrac{x+2}{2x+3}-1+\\dfrac{x+3}{2x+4}-1 \\\\ & \\Leftrightarrow \\dfrac{x-(2x+1)}{2x+1}+\\dfrac{(x+1)-(2x+2)}{2x+2}-\\dfrac{(x+2)-(2x+3)}{2x+3}-\\dfrac{(x+3)-(2x-4)}{2x+4}=0 \\\\ & \\Leftrightarrow \\dfrac{-x-1}{2x+1}+\\dfrac{-x-1}{2x+2}-\\dfrac{-x-1}{2x+3}-\\dfrac{-x-1}{2x+4}=0 \\\\ & \\Leftrightarrow (-x-1)\\left( \\dfrac{1}{2x+1}+\\dfrac{1}{2x+2}-\\dfrac{1}{2x+3}-\\dfrac{1}{2x+4} \\right)=0\\,\\,(*)\\\\ \\end{align} $ <br\/>V\u00ec $ \\dfrac{1}{2x+1}+\\dfrac{1}{2x+2}-\\dfrac{1}{2x+3}-\\dfrac{1}{2x+4}\\ne 0 \\,\\forall \\, x$<br\/> (*) $\\Leftrightarrow -x-1=0 \\\\ \\Leftrightarrow x=-1\\,\\,\\,\\,\\,(\\text{lo\u1ea1i})$ <br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $ S=\\varnothing$<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai<\/span><br\/><span class='basic_green'>Ghi nh\u1edb: <\/span> Ph\u01b0\u01a1ng tr\u00ecnh t\u1ed5ng qu\u00e1t d\u1ea1ng: $\\dfrac{f_1(x)}{g_1(x)}+\\dfrac{f_2(x)}{g_2(x)}+\\dfrac{f_3(x)}{g_3(x)}+...+\\dfrac{f_n(x)}{g_n(x)}=0$ <br\/>+) N\u1ebfu $f_1(x)-g_1(x)\\equiv f_2(x)-g_2(x)\\equiv f_3(x)-g_3(x)\\equiv ...\\equiv f_n(x) - g_n(x)$ th\u00ec ta th\u1ef1c hi\u1ec7n gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh b\u1eb1ng c\u00e1ch tr\u1eeb m\u1ed7i ph\u00e2n th\u1ee9c cho $1$ r\u1ed3i quy \u0111\u1ed3ng v\u00e0 \u0111\u1eb7t nh\u00e2n t\u1eed chung.<\/span>","column":2}]}],"id_ques":1035},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[3]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai22/lv3/img\/11.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{2}{{{x}^{2}}+3x+2}+\\dfrac{2}{{{x}^{2}}+5x+6}+...+\\dfrac{2}{x^2+199x+9900}=\\dfrac{99}{101}.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{1\\}$","B. $S=\\{-102\\}$","C. $S=\\{1;-102\\}$","D. $S=\\varnothing$"],"hint":"Ph\u00e2n t\u00edch m\u1ed7i m\u1eabu th\u1ee9c th\u00e0nh nh\u00e2n t\u1eed r\u1ed3i \u00e1p d\u1ee5ng: $\\dfrac{a}{x(x+a)}=\\dfrac {1}{x}-\\dfrac {1}{x+a}$","explain":" <span class='basic_left'><span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne \\{-100;-99;...;-2;-1\\}$<br\/>Ta c\u00f3:<br\/>$\\dfrac{2}{{{x}^{2}}+3x+2}+\\dfrac{2}{{{x}^{2}}+5x+6}+...+\\dfrac{2}{x^2+199x+9900}=\\dfrac{99}{101}\\\\ \\Leftrightarrow 2\\left[\\dfrac{1}{(x+1)(x+2)}+\\dfrac{1}{(x+2)(x+3)}+...+\\dfrac{1}{(x+99)(x+100)}\\right]=\\dfrac{99}{101}\\\\ \\Leftrightarrow 2\\left(\\dfrac {1}{x+1}-\\dfrac {1}{x+2}+\\dfrac {1}{x+2}-\\dfrac {1}{x+3}+...+\\dfrac{1}{x+99}-\\dfrac {1}{x+100}\\right)=\\dfrac {99}{101}\\\\ \\Leftrightarrow \\dfrac {1}{x+1}-\\dfrac {1}{x+100}=\\dfrac {99}{202}\\\\ \\Leftrightarrow 202(x+100)-202(x+1)=99(x+1)(x+100)\\\\ \\Leftrightarrow 19998=99(x^2+101x+100)\\\\ \\Leftrightarrow x^2+101x+100=202\\\\ \\Leftrightarrow x^2+101x-102=0\\\\ \\Leftrightarrow x^2-x+102x-102=0\\\\ \\Leftrightarrow x(x-1)+102(x-1)=0\\\\ \\Leftrightarrow (x-1)(x+102)=0\\\\ \\Leftrightarrow \\left[\\begin{align}&x=1\\,\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ &x=-102\\,\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ \\end{align}\\right.$<br\/>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{1;-102\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C <\/span><\/span><\/span>","column":2}]}],"id_ques":1036},{"time":24,"part":[{"title":"\u0110i\u1ec1n s\u1ed1 th\u00edch h\u1ee3p v\u00e0o ch\u1ed7 tr\u1ed1ng","title_trans":"","temp":"fill_the_blank","correct":[[["1"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai22/lv3/img\/13.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{10x}{{{x}^{2}}+3x+1}+\\dfrac{-4x}{{{x}^{2}}+2x+1}=1.$<br\/>T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=${_input_}","hint":"Chia c\u1ea3 t\u1eed v\u00e0 m\u1eabu c\u1ee7a m\u1ed7i ph\u00e2n th\u1ee9c cho $x$ r\u1ed3i \u0111\u1eb7t \u1ea9n ph\u1ee5","explain":"<span class='basic_left'><br\/><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Nh\u1eadn x\u00e9t $x=0$ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. Chia c\u1ea3 t\u1eed v\u00e0 m\u1eabu c\u1ee7a hai ph\u00e2n th\u1ee9c cho $x$<br\/><b>B\u01b0\u1edbc 2:<\/b> \u0110\u1eb7t $x+\\dfrac {1}{x}=t$<br\/><b>B\u01b0\u1edbc 3:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$<br\/><b>B\u01b0\u1edbc 4:<\/b> Thay v\u00e0o bi\u1ec3u th\u1ee9c \u1edf b\u01b0\u1edbc 2, gi\u1ea3i t\u00ecm $x$<br\/><b>B\u01b0\u1edbc 5:<\/b> So s\u00e1nh nghi\u1ec7m v\u1edbi \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh v\u00e0 k\u1ebft lu\u1eadn<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x \\ne -1 $<br\/>V\u00ec $ x = 0 $ kh\u00f4ng l\u00e0 nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh n\u00ean ta c\u00f3:<br\/>$\\begin{align}& \\dfrac{10x}{{{x}^{2}}+3x+1}+\\dfrac{-4x}{{{x}^{2}}+2x+1}=1 \\\\ & \\Leftrightarrow \\dfrac{10}{x+3+\\dfrac{1}{x}}+\\dfrac{-4}{x+2+\\dfrac{1}{x}}=1 \\\\ \\end{align}$ (Chia c\u1ea3 t\u1eed v\u00e0 m\u1eabu c\u1ee7a m\u1ed7i ph\u00e2n th\u1ee9c cho $x$)<br\/>\u0110\u1eb7t $t=x+\\dfrac {1}{x}$. Ta c\u00f3: <br\/>$ \\dfrac {10}{t+3}+\\dfrac {-4}{t+2}=1\\,\\,\\,( t \\ne -3 \\text{ v\u00e0 } t \\ne -2)\\\\ \\Leftrightarrow \\dfrac {10(t+2)}{(t+3)(t+2)}+\\dfrac {-4(t+3)}{(t+2)(t+3)}=\\dfrac {(t+2)(t+3)}{(t+2)(t+3)}\\\\ \\Rightarrow 10t+20-4t-12=t^2+5t+6\\\\ \\Leftrightarrow t^2-t-2=0\\\\ \\Leftrightarrow t^2+t-2t-2=0\\\\ \\Leftrightarrow (t+1)(t-2)=0\\\\ \\Leftrightarrow \\left[\\begin {aligned}& t=-1\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ & t=2\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ \\end{aligned}\\right.$<br\/>V\u1edbi $ t = 2 $, ta c\u00f3: <br\/>$x+\\dfrac {1}{x}=2\\\\ \\Rightarrow x^2-2x+1=0\\\\ \\Leftrightarrow (x-1)^2=0\\\\ \\Leftrightarrow x=1\\,\\,\\text{(th\u1ecfa m\u00e3n)}$<br\/>V\u1edbi $ t=-1$, ta c\u00f3:<br\/>$ x+\\dfrac {1}{x}=-1\\\\ \\Rightarrow x^2+x+1=0\\\\ \\Leftrightarrow \\left(x+\\dfrac{1}{2}\\right)^2=\\dfrac{-3}{4}\\,\\,\\text{(v\u00f4 nghi\u1ec7m)}$<br\/><span class='basic_pink'>V\u1eady t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 $S=\\{1\\}$<\/span><\/span>"}]}],"id_ques":1037},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0110\u00fang ho\u1eb7c Sai","title_trans":"","temp":"multiple_choice","correct":[[2]],"list":[{"point":10,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai22/lv3/img\/12.jpg' \/><\/center>Ph\u01b0\u01a1ng tr\u00ecnh $(2x+7)(x+3)^2(2x+5)=18$ c\u00f3 $3$ nghi\u1ec7m.","select":["\u0110\u00fang","Sai"],"hint":"Nh\u00e2n hai v\u1ebf c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi $4$. V\u1ebf tr\u00e1i nh\u00e2n nh\u00e2n t\u1eed th\u1ee9 hai v\u1edbi $4$. R\u1ed3i \u0111\u1eb7t \u1ea9n ph\u1ee5.","explain":" <span class='basic_left'><br\/><span class='basic_green'>H\u01b0\u1edbng d\u1eabn<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> Nh\u00e2n hai v\u1ebf c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi $4$. \u0110\u01b0a ph\u01b0\u01a1ng tr\u00ecnh v\u1ec1 d\u1ea1ng $(2x+7)(2x+6)^2(2x+5)=72$<br\/><b>B\u01b0\u1edbc 2:<\/b> \u0110\u1eb7t $2x+6=t$<br\/><b>B\u01b0\u1edbc 3:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh \u1ea9n $t$<br\/><b>B\u01b0\u1edbc 4:<\/b> Thay v\u00e0o bi\u1ec3u th\u1ee9c \u1edf b\u01b0\u1edbc 2, gi\u1ea3i t\u00ecm $x$<br\/><b>B\u01b0\u1edbc 5:<\/b> K\u1ebft lu\u1eadn.<br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>Ta c\u00f3:<br\/>$(2x+7)(x+3)^2(2x+5)=18\\\\ \\Leftrightarrow (2x+7)[4(x+3)^2](2x+5)=72\\\\ \\Leftrightarrow (2x+7)(2x+6)^2(2x+5)=72$<br\/>\u0110\u1eb7t $2x+6=t\\,\\Rightarrow 2x+7=t+1; 2x+5=t-1$<br\/>Ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh:<br\/>$(t+1).t^2.(t-1)=72\\\\ \\Leftrightarrow t^2(t^2-1)=72\\\\ \\Leftrightarrow t^4-t^2-72=0\\\\ \\Leftrightarrow t^4+8t^2-9t^2-72=0\\\\ \\Leftrightarrow (t^2+8)(t^2-9)=0\\\\ \\Leftrightarrow \\left[\\begin{align}&t^2=-8\\,\\,\\text{(lo\u1ea1i)}\\\\&t^2=9\\\\ \\end{align}\\right. \\\\ \\Leftrightarrow \\left[\\begin{aligned}&t=-3\\\\ &t=3\\\\ \\end{aligned}\\right.$ <br\/>V\u1edbi $t=3$, ta c\u00f3:<br\/>$2x+6=3\\Leftrightarrow x=\\dfrac {-3}{2}$<br\/>V\u1edbi $t=-3$, ta c\u00f3:<br\/>$2x+6=-3\\Leftrightarrow x=\\dfrac {-9}{2}$ <br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho c\u00f3 hai nghi\u1ec7m.<br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai.<\/span><\/span>","column":2}]}],"id_ques":1038},{"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":[[["420"],["300"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"<span class='basic_left'>Trong th\u00e1ng $1$, hai t\u1ed5 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c $720$ s\u1ea3n ph\u1ea9m. Trong th\u00e1ng $2$, t\u1ed5 $1$ v\u01b0\u1ee3t m\u1ee9c $15\\%$ so v\u1edbi th\u00e1ng $1$; t\u1ed5 $2$ v\u01b0\u1ee3t m\u1ee9c $12\\%$ so v\u1edbi th\u00e1ng $1$ n\u00ean s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c $819$ s\u1ea3n ph\u1ea9m. H\u1ecfi trong th\u00e1ng $1$, m\u1ed7i t\u1ed5 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c bao nhi\u00eau s\u1ea3n ph\u1ea9m.<br\/><b> \u0110\u00e1p s\u1ed1:<\/b> T\u1ed5 $1$: _input_ (s\u1ea3n ph\u1ea9m); T\u1ed5 $2$: _input_(s\u1ea3n ph\u1ea9m)<\/span>","hint":"G\u1ecdi s\u1ed1 s\u1ea3n ph\u1ea9m t\u1ed5 $1$ l\u00e0m \u0111\u01b0\u1ee3c trong th\u00e1ng $1$ l\u00e0 \u1ea9n s\u1ed1. Bi\u1ec3u di\u1ec5n s\u1ed1 s\u1ea3n ph\u1ea9m t\u1ed5 $2$ l\u00e0m \u0111\u01b0\u1ee3c theo \u1ea9n s\u1ed1.","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> G\u1ecdi s\u1ed1 s\u1ea3n ph\u1ea9m t\u1ed5 $1$ l\u00e0m \u0111\u01b0\u1ee3c trong th\u00e1ng $1$ l\u00e0 \u1ea9n s\u1ed1<br\/><b>B\u01b0\u1edbc 2:<\/b> Bi\u1ec3u di\u1ec5n s\u1ed1 s\u1ea3n ph\u1ea9m t\u1ed5 $2$ l\u00e0m \u0111\u01b0\u1ee3c theo \u1ea9n<br\/><b>B\u01b0\u1edbc 3:<\/b> Bi\u1ec3u di\u1ec5n s\u1ed1 s\u1ea3n ph\u1ea9m hai t\u1ed5 l\u00e0m \u0111\u01b0\u1ee3c trong th\u00e1ng $2$.<br\/><b>B\u01b0\u1edbc 4:<\/b> L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh, gi\u1ea3i v\u00e0 k\u1ebft lu\u1eadn.<br\/>L\u1eadp b\u1ea3ng<br\/><table><tr><th><\/th><th>T\u1ed5 $1$<\/th><th>T\u1ed5 $2$<\/th><th>T\u1ed5ng<\/th><\/tr><tr><td>Th\u00e1ng $1$<\/td><td>$x$<\/td><td>$720-x$<\/td><td>$720$<\/td><\/tr><tr><td>Th\u00e1ng $2$<\/td><td>$x+15\\%x$<\/td><td>$(720-x)+12\\%(720-x)$<\/td><td>$819$<\/td><\/tr><\/table><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi s\u1ed1 s\u1ea3n ph\u1ea9m t\u1ed5 $1$ l\u00e0m \u0111\u01b0\u1ee3c trong th\u00e1ng $1$ l\u00e0 $x$ (s\u1ea3n ph\u1ea9m, $ 0 < x < 720 $)<br\/>V\u00ec trong th\u00e1ng $1$, hai t\u1ed5 l\u00e0m \u0111\u01b0\u1ee3c $720$ s\u1ea3n ph\u1ea9m, n\u00ean t\u1ed5 $2$ l\u00e0m \u0111\u01b0\u1ee3c $720-x$ (s\u1ea3n ph\u1ea9m)<br\/>Th\u00e1ng $2$, t\u1ed5 $1$ v\u01b0\u1ee3t m\u1ee9c $15\\%$ so v\u1edbi th\u00e1ng $1$, n\u00ean t\u1ed5 $1$ l\u00e0m \u0111\u01b0\u1ee3c l\u00e0 $x+15\\%x=1,15 x$ (s\u1ea3n ph\u1ea9m)<br\/>Th\u00e1ng $2$, t\u1ed5 $2$ l\u00e0m v\u01b0\u1ee3t m\u1ee9c $12\\%$ so v\u1edbi th\u00e1ng $1$, n\u00ean t\u1ed5 $2$ l\u00e0m \u0111\u01b0\u1ee3c l\u00e0 $(720-x)+12\\%(720-x)=806,4-1,12x$ (s\u1ea3n ph\u1ea9m)<br\/>V\u00ec th\u00e1ng $2$, hai t\u1ed5 l\u00e0m \u0111\u01b0\u1ee3c $819 $ s\u1ea3n ph\u1ea9m n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$1,15x+806,4+x-1,12x=819\\\\ \\Leftrightarrow 0,03x=12,6\\\\ \\Leftrightarrow x=420\\,\\,\\text{(th\u1ecfa m\u00e3n)}$<br\/>V\u1eady th\u00e1ng $1$ t\u1ed5 $1$ l\u00e0m \u0111\u01b0\u1ee3c $420$ s\u1ea3n ph\u1ea9m; t\u1ed5 $2$ l\u00e0m \u0111\u01b0\u1ee3c $720-420=300$ (s\u1ea3n ph\u1ea9m)<br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $420$ v\u00e0 $300$<\/span><\/span>"}]}],"id_ques":1039},{"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":[[["20"]]],"list":[{"point":10,"width":50,"type_input":"","ques":"M\u1ed9t chi\u1ebfc thuy\u1ec1n kh\u1edfi h\u00e0nh t\u1eeb b\u1ebfn s\u00f4ng $A$. Sau $45$ ph\u00fat, m\u1ed9t ca n\u00f4 t\u1eeb b\u1ebfn s\u00f4ng $A$ \u0111u\u1ed5i theo v\u00e0 g\u1eb7p thuy\u1ec1n c\u00e1ch b\u1ebfn $A$ l\u00e0 $40 km$. Bi\u1ebft v\u1eadn t\u1ed1c c\u1ee7a ca n\u00f4 h\u01a1n v\u1eadn t\u1ed1c c\u1ee7a thuy\u1ec1n l\u00e0 $12 km\/h$. T\u00ednh v\u1eadn t\u1ed1c thuy\u1ec1n.(Gi\u1ea3 s\u1eed ca n\u00f4 v\u00e0 thuy\u1ec1n \u0111i tr\u00ean d\u00f2ng s\u00f4ng t\u0129nh l\u1eb7ng.)<br\/><b> \u0110\u00e1p s\u1ed1: <\/b> _input_ ($km\/h$)","hint":"D\u00f2ng s\u00f4ng t\u0129nh l\u1eb7ng c\u00f3 v\u1eadn t\u1ed1c d\u00f2ng n\u01b0\u1edbc l\u00e0 $0\\,km\/h$. <br\/>Qu\u00e3ng s\u00f4ng thuy\u1ec1n v\u00e0 ca n\u00f4 \u0111i b\u1eb1ng nhau v\u00e0 b\u1eb1ng $40km$","explain":"<span class='basic_left'><span class='basic_green'>H\u01b0\u1edbng d\u1eabn:<\/span><br\/><b>B\u01b0\u1edbc 1:<\/b> G\u1ecdi v\u1eadn t\u1ed1c thuy\u1ec1n l\u00e0 \u1ea9n s\u1ed1<br\/><b>B\u01b0\u1edbc 2:<\/b> Bi\u1ec3u di\u1ec5n v\u1eadn t\u1ed1c ca n\u00f4 theo \u1ea9n<br\/><b>B\u01b0\u1edbc 3:<\/b> Bi\u1ec3u di\u1ec5n th\u1eddi gian c\u1ee7a thuy\u1ec1n v\u00e0 ca n\u00f4<br\/><b>B\u01b0\u1edbc 4:<\/b> L\u1eadp ph\u01b0\u01a1ng tr\u00ecnh d\u1ef1a tr\u00ean d\u1eef ki\u1ec7n ca n\u00f4 \u0111i sau thuy\u1ec1n $45$ ph\u00fat.<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>Thuy\u1ec1n<\/td><td>$40$<\/td><td>$x$<\/td><td>$\\dfrac{40}{x}$<\/td><\/tr><tr><td>Ca n\u00f4<\/td><td>$40$<\/td><td>$x+12$<\/td><td>$\\dfrac{40}{x+12}$<\/td><\/tr><\/table><br\/><span class='basic_green'>B\u00e0i gi\u1ea3i<\/span><br\/>G\u1ecdi v\u1eadn t\u1ed1c c\u1ee7a thuy\u1ec1n l\u00e0 $x$ ($km\/h$, $x>0$)<br\/>V\u00ec v\u1eadn t\u1ed1c ca n\u00f4 l\u1edbn h\u01a1n v\u1eadn t\u1ed1c c\u1ee7a thuy\u1ec1n l\u00e0 $12km\/h $ n\u00ean v\u1eadn t\u1ed1c ca n\u00f4 l\u00e0 $x+12$ ($km\/h$)<br\/>V\u00ec khi ca n\u00f4 g\u1eb7p thuy\u1ec1n c\u00e1ch b\u1ebfn $A$ l\u00e0 $40km$ n\u00ean thuy\u1ec1n v\u00e0 ca n\u00f4 c\u00f9ng \u0111i \u0111\u01b0\u1ee3c qu\u00e3ng s\u00f4ng l\u00e0 $40km$<br\/>Do v\u1eady, th\u1eddi gian thuy\u1ec1n \u0111i \u0111\u1ebfn khi g\u1eb7p ca n\u00f4 l\u00e0 $\\dfrac {40}{x}$ (gi\u1edd)<br\/>Th\u1eddi gian ca n\u00f4 \u0111i \u0111\u1ebfn khi g\u1eb7p thuy\u1ec1n l\u00e0 $\\dfrac {40}{x+12}$(gi\u1edd)<br\/>V\u00ec ca n\u00f4 \u0111i sau thuy\u1ec1n $45$ ph\u00fat $=\\dfrac {3}{4}$ gi\u1edd n\u00ean ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh:<br\/>$\\dfrac {40}{x}-\\dfrac {40}{x+12}=\\dfrac {3}{4}\\\\ \\Rightarrow 160(x+12)-160x=3x(x+12)\\\\ \\Leftrightarrow 3x^2+36x-1920=0\\\\ \\Leftrightarrow (x-20)(x+32)=0\\\\ \\Leftrightarrow \\left[\\begin{align} &x=20\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ &x=-32\\,\\,\\text{(lo\u1ea1i)}\\\\ \\end{align}\\right.$<br\/>V\u1eady v\u1eadn t\u1ed1c c\u1ee7a thuy\u1ec1n l\u00e0 $20\\,km\/h$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $20$<\/span><\/span>"}]}],"id_ques":1040}],"lesson":{"save":0,"level":3}}