{"segment":[{"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/bai20/lv1/img\/12.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu th\u1ee9c l\u00e0 gi\u00e1 tr\u1ecb c\u1ee7a \u1ea9n \u0111\u1ec3 t\u1ea5t c\u1ea3 c\u00e1c m\u1eabu th\u1ee9c trong ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec1u kh\u00e1c $0$.","select":["\u0110\u00fang","Sai"],"hint":"","explain":" \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu th\u1ee9c l\u00e0 gi\u00e1 tr\u1ecb c\u1ee7a \u1ea9n \u0111\u1ec3 t\u1ea5t c\u1ea3 c\u00e1c m\u1eabu th\u1ee9c trong ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ec1u kh\u00e1c $0$.<br\/><span class='basic_pink'>Kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 \u0111\u00fang.<\/span>","column":2}]}],"id_ques":891},{"time":24,"part":[{"time":3,"title":"N\u1ed1i m\u1ed7i ph\u01b0\u01a1ng tr\u00ecnh c\u1ed9t tr\u00e1i v\u1edbi \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh t\u01b0\u01a1ng \u1ee9ng \u1edf c\u1ed9t ph\u1ea3i. ","title_trans":"","audio":"","temp":"matching","correct":[["2","4","1","3"]],"list":[{"point":5,"image":"","left":["$\\dfrac{1}{3x-2}=4$","$\\dfrac{1}{{{x}^{2}}-9}=\\dfrac{7x-1}{x+3}$","$\\dfrac{x}{x-1}+\\dfrac{1}{x+3}=1$","$\\dfrac{1}{{{x}^{2}}+1}=2$"],"right":["$x\\ne 1$ v\u00e0 $x\\ne -3$","$x\\ne \\dfrac{2}{3}$","$x\\in \\mathbb R.$","$x\\ne 3$ v\u00e0 $x\\ne -3$"],"top":100,"hint":"","explain":"<span class='basic_left'>+) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $\\dfrac{1}{3x-2}=4$ l\u00e0 $3x-2\\ne 0 \\Leftrightarrow x\\ne \\dfrac {2}{3}$<br\/>+) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $\\dfrac{1}{{{x}^{2}}-9}=\\dfrac{7x-1}{x+3}$ l\u00e0:<br\/>$\\left\\{ \\begin{aligned} & {{x}^{2}}-9\\ne 0 \\\\ & x+3\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & (x-3)(x+3)\\ne 0 \\\\ & x\\ne -3 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne 3 \\\\ & x\\ne -3 \\\\ \\end{aligned} \\right.$ <br\/>+) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $\\dfrac{x}{x-1}+\\dfrac{1}{x+3}=1$ l\u00e0: <br\/>$\\left\\{ \\begin{aligned} & x-1\\ne 0 \\\\ & x+3\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned}& x\\ne 1 \\\\ & x\\ne -3 \\\\ \\end{aligned} \\right.$. <br\/>+) \u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a $\\dfrac{1}{{{x}^{2}}+1}=2$ l\u00e0 $x^2+1\\ne 0 \\Leftrightarrow x^2 \\ne -1$ lu\u00f4n \u0111\u00fang v\u1edbi m\u1ecdi $x \\in \\mathbb R$. (V\u00ec $x^2\\ge 0\\,\\forall x\\in\\mathbb R$)<br\/><\/span>"}]}],"id_ques":892},{"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/bai20/lv1/img\/10.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{3-x}{x(x-2)}+\\dfrac{3}{x}=1$ l\u00e0:","select":["A. $ x\\ne 0.$","B. $x\\ne 0$ v\u00e0 $x\\ne 2$","C. $x\\ne0$ v\u00e0 $x\\ne 3$ v\u00e0 $x\\ne 2$","D. $x\\ne 0$ ho\u1eb7c $x\\ne 2$"],"hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{3-x}{x(x-2)}+\\dfrac{3}{x}=1$ l\u00e0:<br\/>$\\left\\{\\begin{aligned}&x\\ne 0 \\\\ &x(x-2) \\ne 0 \\\\ \\end {aligned}\\right.\\\\ \\Leftrightarrow \\left\\{\\begin{aligned}&x\\ne 0 \\\\ &x-2 \\ne 0 \\\\ \\end {aligned}\\right. \\\\ \\Leftrightarrow \\left \\{\\begin{aligned} & x\\ne 0 \\\\ &x \\ne 2 \\\\ \\end{aligned}\\right.$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 B<\/span><\/span>","column":2}]}],"id_ques":893},{"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/bai20/lv1/img\/10.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x}{{{x}^{2}}-3x}+\\dfrac{1}{{{\\left( x-3 \\right)}^{2}}}=2$ l\u00e0:","select":["A. $x\\ne 3$","B. $x\\ne 0$","C. $x\\ne0$ v\u00e0 $x\\ne 3$","D. $x\\ne 3$ ho\u1eb7c $x\\ne 0$"],"hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x}{{{x}^{2}}-3x}+\\dfrac{1}{{{\\left( x-3 \\right)}^{2}}}=2$ l\u00e0:<br\/>$\\left\\{ \\begin{aligned} & {{x}^{2}}-3x\\ne 0 \\\\ & {{\\left( x-3 \\right)}^{2}}\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x(x-3)\\ne 0\\\\ & x-3\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne 0 \\\\ & x\\ne 3 \\\\ \\end{aligned} \\right.$ <br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 C<\/span><\/span>","column":2}]}],"id_ques":894},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"fill_the_blank_random","correct":[[["-2"],["2"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/13.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{x-2}+\\dfrac{2}{x+2}=\\dfrac{3}{{{x}^{2}}-4}$ l\u00e0: <br\/>$x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$ v\u00e0 $x\\ne \\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$","hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{x-2}+\\dfrac{2}{x+2}=\\dfrac{3}{{{x}^{2}}-4}$ l\u00e0:<br\/>$\\left\\{ \\begin{aligned} & x-2\\ne 0 \\\\ & x+2\\ne 0 \\\\ & {{x}^{2}}-4\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne 2 \\\\ & x\\ne -2 \\\\ & (x-2)(x+2)\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & x\\ne 2 \\\\ & x\\ne -2 \\\\ \\end{aligned} \\right.$ <br\/><span class='basic_pink'>V\u1eady c\u00e1c s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $2$ v\u00e0 $-2$<\/span><\/span>"}]}],"id_ques":895},{"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/bai20/lv1/img\/14.jpg' \/><\/center>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{{{x}^{2}}+1}+\\dfrac{2}{{{x}^{2}}+2}=3$ l\u00e0 $x\\ne 1$ v\u00e0 $x\\ne 2$. <br\/><b> \u0110\u00fang<\/b> hay <b> Sai<\/b>?","select":["\u0110\u00fang","Sai"],"hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{{{x}^{2}}+1}+\\dfrac{2}{{{x}^{2}}+2}=3$ l\u00e0:<br\/>$\\left\\{ \\begin{aligned} & {{x}^{2}}+1\\ne 0 \\\\ & {{x}^{2}}+2\\ne 0 \\\\ \\end{aligned} \\right.\\Leftrightarrow \\left\\{ \\begin{aligned} & {{x}^{2}}\\ne -1 \\\\ & {{x}^{2}}\\ne -2 \\\\ \\end{aligned} \\right.\\,\\,\\,\\,\\,\\,\\,\\,\\,\\forall x\\in \\mathbb{R}\\, (\\text{V\u00ec } x^2\\ge 0\\,\\forall x)$ <br\/><span class='basic_pink'>V\u1eady kh\u1eb3ng \u0111\u1ecbnh tr\u00ean l\u00e0 Sai.<\/span><\/span>","column":2}]}],"id_ques":896},{"time":24,"part":[{"time":3,"title":"S\u1eafp x\u1ebfp c\u00e1c c\u00e2u sau theo th\u1ee9 t\u1ef1 c\u00e1c b\u01b0\u1edbc gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu.","title_trans":"","temp":"sequence","correct":[[[3],[1],[2],[4]]],"list":[{"point":5,"image":"img\/1.png","left":["Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1eeba nh\u1eadn \u0111\u01b0\u1ee3c.","T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh","Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c r\u1ed3i kh\u1eed m\u1eabu","Lo\u1ea1i c\u00e1c gi\u00e1 tr\u1ecb t\u00ecm \u0111\u01b0\u1ee3c kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh r\u1ed3i k\u1ebft lu\u1eadn."],"top":55,"hint":"","explain":"<span class='basic_left'>C\u00e1c b\u01b0\u1edbc gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n m\u1eabu th\u1ee9c:<br\/><b>B\u01b0\u1edbc 1:<\/b> T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<br\/><b>B\u01b0\u1edbc 2:<\/b> Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c c\u00e1c ph\u00e2n th\u1ee9c r\u1ed3i kh\u1eed m\u1eabu.<br\/><b>B\u01b0\u1edbc 3:<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1eeba nh\u1eadn \u0111\u01b0\u1ee3c.<br\/><b>B\u01b0\u1edbc 4:<\/b> Lo\u1ea1i c\u00e1c gi\u00e1 tr\u1ecb t\u00ecm \u0111\u01b0\u1ee3c kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh r\u1ed3i k\u1ebft lu\u1eadn. <\/span>"}]}],"id_ques":897},{"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":[[["0"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/12.jpg' \/><center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{x-2}+3=\\dfrac{2x-5}{x-2}.$<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":"Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c r\u1ed3i kh\u1eed m\u1eabu \u0111\u1ec3 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 2$.<br\/>Ta c\u00f3:<br\/>$\\begin{aligned}& \\dfrac{1}{x-2}+3=\\dfrac{2x-5}{x-2} \\\\ & \\Leftrightarrow \\dfrac{1}{x-2}+\\dfrac{3\\left( x-2 \\right)}{x-2}=\\dfrac{2x-5}{x-2} \\\\ & \\Rightarrow 1+3\\left( x-2\\right)=2x-5 \\\\ & \\Leftrightarrow 1+3x-6-2x+5=0 \\\\ & \\Leftrightarrow x=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n}) \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{0\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $0$<\/span><br\/><i><span class='basic_green'>Ch\u00fa \u00fd:<\/span><\/i><br\/>- Khi gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu th\u1ee9c, b\u01b0\u1edbc kh\u1eed m\u1eabu s\u1eed d\u1ee5ng d\u1ea5u \u201c$\\Rightarrow$\u201d thay v\u00ec d\u1ea5u \u201c$\\Leftrightarrow$\u201d v\u00ec t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh sau khi kh\u1eed m\u1eabu c\u00f3 th\u1ec3 ch\u1ee9a t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u ch\u1ee9 kh\u00f4ng b\u1eb1ng t\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh ban \u0111\u1ea7u do v\u1eady hai ph\u01b0\u01a1ng tr\u00ecnh kh\u00f4ng t\u01b0\u01a1ng \u0111\u01b0\u01a1ng.<br\/><b>V\u00ed d\u1ee5:<\/b> Ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac {x^2-4}{x-2}=0$ ch\u1ec9 c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{2\\}$. <br\/>Nh\u01b0ng khi kh\u1eed m\u1eabu ph\u01b0\u01a1ng tr\u00ecnh tr\u1edf th\u00e0nh $x^2-4=0$ c\u00f3 t\u1eadp nghi\u1ec7m l\u00e0 $S=\\{-2;2\\}$<br\/>Ta g\u1ecdi ph\u01b0\u01a1ng tr\u00ecnh sau khi kh\u1eed m\u1eabu l\u00e0 <i>ph\u01b0\u01a1ng tr\u00ecnh h\u1ec7 qu\u1ea3 <\/i> c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho.<\/span>"}]}],"id_ques":898},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/9.jpg' \/><\/center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x}{x-2}=\\dfrac{x-3}{x-4}.$<br\/> T\u1eadp nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0:","select":["A. $S=\\{6\\}$","B. $S=\\mathbb R$","C. $S=\\varnothing$","D. $S=\\{-6\\}$"],"hint":"","explain":" <span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 2$ v\u00e0 $x\\ne 4$.<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & \\dfrac{x}{x-2}=\\dfrac{x-3}{x-4} \\\\ & \\Leftrightarrow \\dfrac{x\\left( x-4 \\right)}{\\left( x-2 \\right)\\left( x-4 \\right)}=\\dfrac{\\left( x-3 \\right)\\left( x-2\\right)}{\\left( x-2 \\right)\\left( x-4 \\right)} \\\\ & \\Rightarrow x(x-4)=(x-3)(x-2) \\\\ &\\Leftrightarrow {{x}^{2}}-4x={{x}^{2}}-5x+6 \\\\ & \\Leftrightarrow 5x-4x=6 \\\\ & \\Leftrightarrow x=6 \\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{6\\}$<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><br\/><i><span class='basic_green'>L\u01b0u \u00fd: <\/span><\/i> Nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 gi\u00e1 tr\u1ecb c\u1ee7a $x$ th\u1ecfa m\u00e3n ph\u01b0\u01a1ng tr\u00ecnh \u0111\u1ed3ng th\u1eddi th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh. V\u00ec v\u1eady, sau khi gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh lu\u00f4n c\u1ea7n lo\u1ea1i c\u00e1c gi\u00e1 tr\u1ecb $x$ kh\u00f4ng th\u1ecfa m\u00e3n \u0111i\u1ec1u ki\u1ec7n ph\u01b0\u01a1ng tr\u00ecnh r\u1ed3i k\u1ebft lu\u1eadn. <\/span>","column":2}]}],"id_ques":899},{"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":5,"width":50,"type_input":"","input_hint":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/8.jpg' \/><center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{12}{1-9{{x}^{2}}}=\\dfrac{1-3x}{1+3x}-\\dfrac{1+3x}{1-3x}.$<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":"Ph\u00e2n t\u00edch m\u1eabu th\u1ee9c \u1edf ph\u00e2n th\u1ee9c th\u1ee9 nh\u1ea5t th\u00e0nh nh\u00e2n t\u1eed. <br\/>Quy \u0111\u1ed3ng m\u1eabu th\u1ee9c r\u1ed3i kh\u1eed m\u1eabu \u0111\u1ec3 gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne \\dfrac 1 3$ v\u00e0 $x\\ne \\dfrac {-1} 3$.<br\/>Ta c\u00f3:<br\/>$\\begin{aligned} & \\dfrac{12}{1-9{{x}^{2}}}=\\dfrac{1-3x}{1+3x}-\\dfrac{1+3x}{1-3x} \\\\ & \\Leftrightarrow \\dfrac{12}{(1-3x)(1+3x)}=\\dfrac{1-3x}{1+3x}-\\dfrac{1+3x}{1-3x} \\\\ & \\Leftrightarrow \\dfrac{12}{(1-3x)(1+3x)}=\\dfrac{(1-3x)(1-3x)}{(1-3x)(1+3x)}-\\dfrac{(1+3x)(1+3x)}{(1-3x)(1+3x)} \\\\ & \\Rightarrow 12={{\\left( 1-3x \\right)}^{2}}-{{\\left( 1+3x \\right)}^{2}} \\\\ & \\Leftrightarrow \\left( 1-6x+9{{x}^{2}} \\right)-\\left( 1+6x+9{{x}^{2}} \\right)-12=0 \\\\ & \\Leftrightarrow 1-6x+9{{x}^{2}}-1-6x-9{{x}^{2}}-12=0 \\\\ & \\Leftrightarrow -12x-12=0 \\\\ & \\Leftrightarrow x=-1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n }) \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-1\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$<\/span><br\/><\/span>"}]}],"id_ques":900},{"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":[[["3"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/7.jpg' \/><center>Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{3x-5}{x-1}-\\dfrac{2x-5}{x-2}=1.$<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 x\u00e1c \u0111\u1ecbnh: $x\\ne 1$ v\u00e0 $x\\ne 2$.<br\/>Ta c\u00f3:<br\/>$\\begin{aligned}& \\dfrac{3x-5}{x-1}-\\dfrac{2x-5}{x-2}=1 \\\\ & \\Leftrightarrow \\dfrac{\\left( 3x-5 \\right)\\left( x-2 \\right)}{\\left( x-1 \\right)\\left( x-2 \\right)}-\\dfrac{\\left( 2x-5 \\right)\\left( x-1 \\right)}{\\left( x-1 \\right)\\left( x-2 \\right)}=\\dfrac{\\left( x-1 \\right)\\left( x-2 \\right)}{\\left( x-1 \\right)\\left( x-2 \\right)} \\\\ & \\Rightarrow \\left( 3x-5 \\right)\\left( x-2 \\right)-\\left( 2x-5 \\right)\\left( x-1 \\right)=\\left( x-1 \\right)\\left( x-2 \\right) \\\\ & \\Leftrightarrow 3{{x}^{2}}-11x+10-\\left( 2{{x}^{2}}-7x+5 \\right)-\\left( {{x}^{2}}-3x+2 \\right)=0 \\\\ & \\Leftrightarrow 3{{x}^{2}}-11x+10-2{{x}^{2}}+7x-5-{{x}^{2}}+3x-2=0 \\\\ & \\Leftrightarrow -x+3=0 \\\\ & \\Leftrightarrow x=3\\,\\,(\\text{th\u1ecfa m\u00e3n)} \\\\ \\end{aligned}$<br\/>V\u1eady ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{3\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $3$<\/span><br\/><\/span>"}]}],"id_ques":901},{"time":24,"part":[{"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":[[["-1"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"","ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x+m}{x-2}-\\dfrac{x-m}{2+x}=\\dfrac{2m}{{{x}^{2}}-4}.$ ($m$ l\u00e0 tham s\u1ed1).<br\/> <b>C\u00e2u 1. <\/b>T\u00ecm $m$ \u0111\u1ec3 ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m l\u00e0 $x=-1$<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $m=\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}$<\/span> ","hint":"Thay gi\u00e1 tr\u1ecb c\u1ee7a $x$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh.","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne \\pm 2$.<br\/>Thay $x=-1$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho. Ta c\u00f3:<br\/>$\\begin{align}& \\dfrac{-1+m}{-1-2}-\\dfrac{-1-m}{2+(-1)}=\\dfrac{2m}{{{(-1)}^{2}}-4} \\\\ & \\Leftrightarrow \\dfrac{-1+m}{-3}-\\dfrac{-1-m}{1}=\\dfrac{2m}{-3} \\\\ & \\Leftrightarrow -1+m+3(-1-m)-2m=0 \\\\ & \\Leftrightarrow -4m=4 \\\\ & \\Leftrightarrow m=-1\\\\ \\end{align}$<br\/>V\u1eady v\u1edbi $m=-1$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=-1$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-1$<\/span><\/span>"}]}],"id_ques":902},{"time":24,"part":[{"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":[[["1"],["2"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"frac","ques":"<span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{x+m}{x-2}-\\dfrac{x-m}{2+x}=\\dfrac{2m}{{{x}^{2}}-4}$ ($m$ l\u00e0 tham s\u1ed1).<br\/> <b>C\u00e2u 2.<\/b> Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi $m=2$<\/span> <br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $S=${<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>}","hint":"Thay gi\u00e1 tr\u1ecb c\u1ee7a $m$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh r\u1ed3i gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh.","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne \\pm 2$.<br\/>Thay $m=2$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho. Ta c\u00f3:<br\/>$\\begin{align}& \\dfrac{x+2}{x-2}-\\dfrac{x-2}{x+2}=\\dfrac{4}{{{x}^{2}}-4} \\\\ & \\Leftrightarrow \\dfrac{{{(x+2)}^{2}}}{(x+2)(x-2)}-\\dfrac{{{(x-2)}^{2}}}{(x+2)(x-2)}=\\dfrac{4}{(x-2)(x+2)} \\\\ & \\Rightarrow {{(x+2)}^{2}}-{{(x-2)}^{2}}=4 \\\\ & \\Leftrightarrow{{x}^{2}}+4x+4-{{x}^{2}}+4x-4-4=0 \\\\ & \\Leftrightarrow 8x-4=0 \\\\ & \\Leftrightarrow x=\\dfrac{1}{2} \\,\\,\\,\\,\\,\\,\\text{(th\u1ecfa m\u00e3n)}\\\\ \\end{align}$<br\/>V\u1eady v\u1edbi $m=2$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 m\u1ed9t nghi\u1ec7m $x=\\dfrac{1}{2}$<\/span>"}]}],"id_ques":903},{"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"],["-3","3"],["3","-3"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/13.jpg' \/><center><span class='basic_left'>Cho ph\u01b0\u01a1ng tr\u00ecnh: $\\dfrac{1+m}{1-x}=1+x$ ($m$ l\u00e0 tham s\u1ed1).<br\/>a. T\u00ecm \u0111i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh.<br\/>b. Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh v\u1edbi $m=-9$<br\/><b>\u0110\u00e1p s\u1ed1:<\/b> $a.\\,x\\ne\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\\\ b. \\,S=\\{\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{};\\FormInput[40][bl_elm_true basic_elm basic_blank basic_blank_part]{}\\}$<\/span> ","hint":"Thay $m=-9$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh r\u1ed3i gi\u1ea3i. ","explain":"<span class='basic_left'>\u0110i\u1ec1u ki\u1ec7n x\u00e1c \u0111\u1ecbnh: $x\\ne 1$ <br\/>Thay $m=-9$ v\u00e0o ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho. Ta c\u00f3:<br\/>$\\begin{aligned}& \\dfrac{1-9}{1-x}=1+x \\\\ & \\Leftrightarrow \\dfrac{-8}{1-x}=\\dfrac{(1+x)(1-x)}{1-x} \\\\ & \\Rightarrow -8=1-{{x}^{2}} \\\\ & \\Leftrightarrow {{x}^{2}}=9 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=3 \\,(\\text{th\u1ecfa m\u00e3n})\\\\ & x=-3\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady v\u1edbi $m=-9$ th\u00ec ph\u01b0\u01a1ng tr\u00ecnh c\u00f3 t\u1eadp nghi\u1ec7m $S=\\{-3;3\\}$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $-3$ v\u00e0 $3$<\/span><br\/><\/span>"}]}],"id_ques":904},{"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","temp":"fill_the_blank","correct":[[["-1"],["4"]]],"list":[{"point":5,"width":50,"type_input":"","input_hint":"frac","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/12.jpg' \/><center><span class='basic_left'>Cho hai bi\u1ec3u th\u1ee9c $A=\\dfrac {x+1}{x}+1$ v\u00e0 $B=\\dfrac {2x-1}{x+1}$. T\u00ecm $x$ \u0111\u1ec3 $A=B$<\/span><br\/> <b>\u0110\u00e1p s\u1ed1: <\/b>$x\\in${<div class=\"frac123\"><div class=\"ts\">_input_<\/div><div class=\"ms\">_input_<\/div><\/div>} ","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A=B$. \u0110\u00e2y l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu th\u1ee9c.","explain":"<span class='basic_left'><br\/>Ta c\u00f3:<br\/>$\\begin{align} & A=B \\\\ & \\Leftrightarrow \\dfrac{x+1}{x}+1=\\dfrac{2x-1}{x+1} \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{\u0110i\u1ec1u ki\u1ec7n: } x\\ne 0;x\\ne-1)\\\\ & \\Leftrightarrow \\dfrac{(x+1)(x+1)}{x(x+1)}+\\dfrac{x(x+1)}{x(x+1)}=\\dfrac{x(2x-1)}{x+1} \\\\ & \\Rightarrow {{(x+1)}^{2}}+x(x+1)=x(2x-1) \\\\ & \\Leftrightarrow {{x}^{2}}+2x+1+{{x}^{2}}+x=2{{x}^{2}}-x \\\\ & \\Leftrightarrow 2{{x}^{2}}+3x+1-2{{x}^{2}}+x=0 \\\\ & \\Leftrightarrow 4x=-1 \\\\ & \\Leftrightarrow x=\\dfrac{-1}{4}\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{align}$<br\/>V\u1eady \u0111\u1ec3 $A=B$ th\u00ec $x=\\dfrac{-1}{4}$<\/span>"}]}],"id_ques":905},{"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":5,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/15.jpg' \/><center><span class='basic_left'>Cho hai bi\u1ec3u th\u1ee9c $A=\\dfrac {x+2}{x}$ v\u00e0 $B=\\dfrac{{{x}^{2}}+5x+4}{{{x}^{2}}+2x}-\\dfrac{x}{x+2}$. T\u00ecm $x$ \u0111\u1ec3 $A=B$<br\/> <b>\u0110\u00e1p s\u1ed1:<\/b> $x=$_input_<\/span> ","hint":"","explain":"<span class='basic_left'><br\/>Ta c\u00f3:<br\/>$\\begin{aligned}&A=B \\\\&\\Leftrightarrow\\dfrac{x+2}{x}=\\dfrac{{{x}^{2}}+5x+4}{{{x}^{2}}+2x}-\\dfrac{x}{x+2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{\u0110i\u1ec1u ki\u1ec7n: } x\\ne 0;x\\ne-2) \\\\ & \\Leftrightarrow \\dfrac{x+2}{x}=\\dfrac{(x+1)(x+4)}{x(x+2)}-\\dfrac{x}{x+2} \\\\ & \\Leftrightarrow \\dfrac{{{(x+2)}^{2}}}{x(x+2)}=\\dfrac{{{x}^{2}}+5x+4}{x(x+2)}-\\dfrac{{{x}^{2}}}{x(x+2)} \\\\ & \\Rightarrow {{x}^{2}}+4x+4={{x}^{2}}+5x+4-{{x}^{2}} \\\\ & \\Leftrightarrow {{x}^{2}}+4x+4-{{x}^{2}}-5x-4+{{x}^{2}}=0 \\\\ & \\Leftrightarrow {{x}^{2}}-x=0 \\\\ & \\Leftrightarrow x(x-1)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=0\\,\\,\\,\\,\\,\\,\\,(\\text{lo\u1ea1i}) \\\\ & x=1\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady \u0111\u1ec3 $A=B$ th\u00ec $x=1$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $1$<\/span><br\/><\/span>"}]}],"id_ques":906},{"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":[[["5"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/11.jpg' \/><center><span class='basic_left'>Cho hai bi\u1ec3u th\u1ee9c $A=\\dfrac{2-x}{x-1}+\\dfrac{x-3}{x+1}$ v\u00e0 $B=\\dfrac{2x}{1-{{x}^{2}}}$. T\u00ecm $x$ \u0111\u1ec3 $A=B$<br\/> <b>\u0110\u00e1p s\u1ed1: <\/b>$x\\in${_input_}<\/span> ","hint":"","explain":"<span class='basic_left'><br\/>Ta c\u00f3:<br\/>$\\begin{align}& A=B \\\\ & \\Leftrightarrow \\dfrac{2-x}{x-1}+\\dfrac{x-3}{x+1}=\\dfrac{2x}{1-{{x}^{2}}}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{\u0110i\u1ec1u ki\u1ec7n: } x\\ne\\pm 1) \\\\ & \\Leftrightarrow \\dfrac{(2-x)(x+1)}{(x-1)(x+1)}+\\dfrac{(x-3)(x-1)}{(x-1)(x+1)}=\\dfrac{-2x}{(x-1)(x+1)} \\\\ & \\Rightarrow (2-x)(x+1)+(x-3)(x-1)=-2x \\\\ & \\Leftrightarrow -{{x}^{2}}+x+2+{{x}^{2}}-4x+3+2x=0 \\\\ & \\Leftrightarrow -x+5=0 \\\\ & \\Leftrightarrow x=5\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{align}$<br\/>V\u1eady \u0111\u1ec3 $A=B$ th\u00ec $x=5$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $5$<\/span><br\/><\/span>"}]}],"id_ques":907},{"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":[[["18"],["15"]]],"list":[{"point":5,"width":50,"type_input":"","ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/10.jpg' \/><center><span class='basic_left'>Cho hai bi\u1ec3u th\u1ee9c $A=\\dfrac{90}{x}$ v\u00e0 $B=\\dfrac{36}{x-6}$. T\u00ecm $x$ \u0111\u1ec3 $A-B=2$<br\/> <b>\u0110\u00e1p s\u1ed1:<\/b> $x\\in${_input_;_input_}<\/span> ","hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $A-B=2$. \u0110\u00e2y l\u00e0 ph\u01b0\u01a1ng tr\u00ecnh d\u1ea1ng ph\u01b0\u01a1ng tr\u00ecnh ch\u1ee9a \u1ea9n \u1edf m\u1eabu th\u1ee9c.","explain":"<span class='basic_left'><br\/>Ta c\u00f3:<br\/>$\\begin{aligned}& A-B=2 \\\\ & \\Leftrightarrow \\dfrac{90}{x}-\\dfrac{36}{x-6}=2\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{\u0110i\u1ec1u ki\u1ec7n: }x\\ne 0;x\\ne 6) \\\\ & \\Leftrightarrow \\dfrac{90(x-6)}{x(x-6)}-\\dfrac{36x}{x(x-6)}=\\dfrac{2x(x-6)}{x(x-6)} \\\\ & \\Rightarrow 90(x-6)-36x-2x(x-6)=0 \\\\ & \\Leftrightarrow 90x-540-36x-2{{x}^{2}}+12x=0 \\\\ & \\Leftrightarrow -2{{x}^{2}}+66x-540=0 \\\\ & \\Leftrightarrow {{x}^{2}}-33x+270=0 \\\\ & \\Leftrightarrow {{x}^{2}}-18x-15x+270=0 \\\\ & \\Leftrightarrow x(x-18)-15(x-18)=0 \\\\ & \\Leftrightarrow (x-18)(x-15)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=18\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ & x=15\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n}) \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady \u0111\u1ec3 $A-B=2$ th\u00ec $x=18$ ho\u1eb7c $x=15$<br\/><span class='basic_pink'>V\u1eady s\u1ed1 c\u1ea7n \u0111i\u1ec1n v\u00e0o \u00f4 tr\u1ed1ng l\u00e0 $15$ v\u00e0 $18$<\/span><br\/><\/span>"}]}],"id_ques":908},{"time":24,"part":[{"title":"L\u1ef1a ch\u1ecdn \u0111\u00e1p \u00e1n \u0111\u00fang nh\u1ea5t","title_trans":"","temp":"multiple_choice","correct":[[1]],"list":[{"point":5,"ques":"<center><img src='https://www.luyenthi123.com/file/luyenthi123/lop8/toan/daiso/bai20/lv1/img\/2.jpg' \/><\/center>T\u00ecm $x$ \u0111\u1ec3 bi\u1ec3u th\u1ee9c $\\dfrac{1}{x}+\\dfrac{1}{x+10}$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng $\\dfrac{1}{12}$. ","select":["A. $x\\in\\{-6;20\\}$","B. $x=120$","C. $x\\in\\varnothing$","D. $x\\in\\{30;4\\}$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{1}{x}+\\dfrac{1}{x+10}=\\dfrac{1}{12}$ \u0111\u1ec3 t\u00ecm $x$","explain":" <span class='basic_left'>Ta c\u00f3: <br\/>$\\begin{aligned}& \\dfrac{1}{x}+\\dfrac{1}{x+10}=\\dfrac{1}{12} \\\\ & \\Leftrightarrow \\dfrac{12(x+10)}{12x(x+10)}+\\dfrac{12x}{12x(x+10)}=\\dfrac{x(x+10)}{12x(x+10)}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{\u0110i\u1ec1u ki\u1ec7n: }x\\ne 0;x\\ne -10) \\\\ & \\Rightarrow 12(x+10)+12x-x(x+10)=0 \\\\ & \\Leftrightarrow 12x+120+12x-{{x}^{2}}-10x=0 \\\\ & \\Leftrightarrow {{x}^{2}}-14x-120=0 \\\\ & \\Leftrightarrow {{x}^{2}}-20x+6x-120=0 \\\\ & \\Leftrightarrow x(x-20)+6(x-20)=0 \\\\ & \\Leftrightarrow (x+6)(x-20)=0 \\\\ & \\Leftrightarrow \\left[ \\begin{aligned} & x=-6\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n)} \\\\ & x=20\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{th\u1ecfa m\u00e3n)} \\\\ \\end{aligned} \\right. \\\\ \\end{aligned}$<br\/>V\u1eady $x=-6$ ho\u1eb7c $x=20$.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 A<\/span><\/span>","column":2}]}],"id_ques":909},{"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/bai20/lv1/img\/3.jpg' \/><\/center>T\u00ecm $x$ \u0111\u1ec3 bi\u1ec3u th\u1ee9c $\\dfrac{x-3}{2-x}$ v\u00e0 bi\u1ec3u th\u1ee9c $\\dfrac{-3x+7}{x-2}$ c\u00f3 gi\u00e1 tr\u1ecb b\u1eb1ng nhau.","select":["A. $x=\\dfrac{5}{2}$","B. $x=2$","C. $x=-2$","D. $x\\in\\varnothing$"],"hint":"Gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh $\\dfrac{x-3}{2-x}=\\dfrac{-3x+7}{x-2}$ \u0111\u1ec3 t\u00ecm $x$","explain":" <span class='basic_left'>Ta c\u00f3: <br\/>$\\begin{align} & \\dfrac{x-3}{2-x}=\\dfrac{-3x+7}{x-2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(\\text{\u0110i\u1ec1u ki\u1ec7n: }x\\ne 2) \\\\ & \\Leftrightarrow \\dfrac{3-x}{x-2}=\\dfrac{-3x+7}{x-2} \\\\ & \\Rightarrow 3-x=-3x+7 \\\\ & \\Leftrightarrow 2x=4 \\\\ & \\Leftrightarrow x=2\\,\\,\\,\\,(\\text{lo\u1ea1i}) \\\\ \\end{align}$<br\/>V\u1eady kh\u00f4ng c\u00f3 gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a $x$ th\u1ecfa m\u00e3n \u0111\u1ec3 hai bi\u1ec3u th\u1ee9c tr\u00ean b\u1eb1ng nhau.<br\/><span class='basic_pink'>V\u1eady \u0111\u00e1p \u00e1n \u0111\u00fang l\u00e0 D<\/span><\/span>","column":2}]}],"id_ques":910}],"lesson":{"save":0,"level":1}}