\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2+3}{1-x^2}=0\)
Những câu hỏi liên quan
Tìm `x >= 0`
\(\dfrac{1}{\sqrt{x}+2}>\dfrac{1}{5}\)
\(\dfrac{2}{\sqrt{x}+3}< \dfrac{1}{2}\)
\(\dfrac{\sqrt{x}}{\sqrt{x}+3}>1\)
\(\dfrac{2\sqrt{x}}{\sqrt{x}+1}< \dfrac{1}{3}\)
\(\dfrac{1}{\sqrt{x}+2}>\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{5}>0\)
\(\Leftrightarrow\dfrac{5}{5\sqrt{x}+10}-\dfrac{\sqrt{x}+2}{5\sqrt{x}+10}>0\)
\(\Leftrightarrow\dfrac{5-\sqrt{x}-2}{5\sqrt{x}+10}>0\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-3\right)}{5\sqrt{x}+10}>0\)
Mà: \(5\sqrt{x}+10\ge10>0\forall x\)
\(\Leftrightarrow\sqrt{x}>3\)
\(\Leftrightarrow x>9\)
_________
\(\dfrac{2}{\sqrt{x}+3}< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{4}{2\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+6}< 0\)
\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\sqrt{x}+6}< 0\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-1\right)}{2\sqrt{x}+6}< 0\)
Mà: \(2\sqrt{x}+6\ge6>0\forall x\)
\(\Leftrightarrow\sqrt{x}-1< 0\)
\(\Leftrightarrow\sqrt{x}< 1\)
\(\Leftrightarrow x< 1\)
\(\Leftrightarrow0\le x\le1\)
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22 tháng 3 2021 lúc 20:54
Giải phương trình:
a) \(\dfrac{x^2-x-6}{x-3}=0\)
b) \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
c) \(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)
d) \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
e) \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)Thể loại truyện
a) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)
Suy ra: x+2=0
hay x=-2(thỏa ĐK)
Vậy: S={-2}
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d)
ĐKXĐ: \(x\notin\left\{1;3\right\}\)
Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)
Suy ra: \(x^2-3x+5x-15=x^2-1-8\)
\(\Leftrightarrow2x-15+9=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3(loại)
Vậy: \(S=\varnothing\)
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29 tháng 1 2022 lúc 11:20
Tìm x biết:a,3dfrac{1}{2}-dfrac{1}{2}xdfrac{2}{3}b,dfrac{1}{3}+dfrac{2}{3}:x-7c,dfrac{1}{3}x+dfrac{2}{5}left(x-1right)0d,left(2x-3right)left(6-2xright)0e,x:dfrac{3}{4}+dfrac{1}{4}-dfrac{2}{3}f,dfrac{-2}{3}-dfrac{1}{3}left(2x-5right)dfrac{3}{2}g,2left|dfrac{1}{2}x-dfrac{1}{3}right|-dfrac{3}{2}dfrac{1}{4}h,dfrac{3}{4}-2.left|2x-dfrac{2}{3}right|2i,left(-0,6x-dfrac{1}{2}right).dfrac{3}{4}-left(-1right)dfrac{1}{3}j,left(3x-1right)left(-dfrac{1}{2}x+5right)0k,dfrac{1}{4}+dfrac{1}{3}:left(2x-1right)-5...
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Tìm x biết:
\(a,3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\)
\(b,\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)
\(c,\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(d,\left(2x-3\right)\left(6-2x\right)=0\)
\(e,x:\dfrac{3}{4}+\dfrac{1}{4}=-\dfrac{2}{3}\)
\(f,\dfrac{-2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\)
\(g,2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)
\(h,\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(i,\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\)
\(j,\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(k,\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(l,\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(m,3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(n,60\%x+\dfrac{2}{3}x=\dfrac{1}{3}.6\dfrac{1}{3}\)
\(p,-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(q,3\left(x-\dfrac{1}{2}\right)-5\left(x+\dfrac{3}{5}\right)=-x+\dfrac{1}{5}\)
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
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Giải phương trình:
a) \(\dfrac{x^2}{2}\) + \(\dfrac{18}{x^2}\) = 13(\(\dfrac{x}{2}\) - \(\dfrac{3}{x}\))
b) x(x - 1) + \(\dfrac{1}{x}\)(\(\dfrac{1}{x}\) - 1) = 0
a. ĐKXĐ:...
\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x^2}{4}+\dfrac{9}{x^2}-3+3\right)=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
\(\Leftrightarrow2\left(\dfrac{x}{2}-\dfrac{3}{x}\right)^2+6=13\left(\dfrac{x}{2}-\dfrac{3}{x}\right)\)
Đặt \(\dfrac{x}{2}-\dfrac{3}{x}=t\Rightarrow2t^2-13t+6=0\Rightarrow\left[{}\begin{matrix}t=6\\t=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{3}{x}=6\\\dfrac{x}{2}-\dfrac{3}{x}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
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b. ĐKXĐ: ...
\(\Leftrightarrow x\left(x-1\right)-\dfrac{x-1}{x^2}=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{x^2}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow x=1\)
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Tìm x, biết:
a) \(\dfrac{2}{3}\)x - \(\dfrac{2}{5}\) = \(\dfrac{1}{2}\)x - \(\dfrac{1}{3}\)
b) \(\dfrac{1}{3}\)x + \(\dfrac{2}{5}\) . (x + 1) = 0
\(a,\dfrac{2}{3}x-\dfrac{2}{5}=\dfrac{1}{2}x-\dfrac{1}{3}\\ \Rightarrow\dfrac{2}{3}x-\dfrac{1}{2}x-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\left(\dfrac{2}{3}-\dfrac{1}{2}\right)-\dfrac{2}{5}=-\dfrac{1}{3}\\ \Rightarrow x\dfrac{1}{6}=-\dfrac{11}{15}\\ \Rightarrow x=-\dfrac{22}{5}\\ b,\dfrac{1}{3}x+\dfrac{2}{5}.\left(x+1\right)=0\\ \Rightarrow\dfrac{1}{3}x+\left(x+1\right)=-\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{2}{5}-\left(x+1\right)\\ \Rightarrow\dfrac{1}{3}x=-\dfrac{7}{5}-x\\ \Rightarrow\dfrac{1}{3}.2x=-\dfrac{7}{5}\\ \Rightarrow2x=-\dfrac{21}{5}\\ \Rightarrow x=-\dfrac{21}{10}.\)
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tìm x e Q
a) \(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
b) \(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
c) \(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
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P(dfrac{x-2}{x+2sqrt{x}}+dfrac{1}{sqrt{x+2}}).dfrac{sqrt{x}+1}{sqrt{x}-1} với x 0 và x≠1 A(dfrac{2}{sqrt{x}-2}+dfrac{3}{2sqrt{x+1}}-dfrac{5sqrt{x}-7}{2x-3sqrt{x}-2}):dfrac{2sqrt{x}+3}{5x-10sqrt{x}} với x 0 và x≠4Adfrac{x+1-2sqrt{x}}{sqrt{x-1}}+dfrac{x+sqrt{x}}{sqrt{x}+1} với x≥0 và x≠1V(dfrac{1}{sqrt{x}+2}+dfrac{1}{sqrt{x}-2})dfrac{sqrt{x}+2}{sqrt{x}} với x0 và x≠4A(dfrac{1}{x-1}+dfrac{3sqrt{x}+5}{xsqrt{x}-x-sqrt{x}+1})(dfrac{left(sqrt{x}+1right)^2}{4sqrt{x}}-1)Pdfrac{15sqrt{x}-11}{x...
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P=(\(\dfrac{x-2}{x+2\sqrt{x}}\)+\(\dfrac{1}{\sqrt{x+2}}\)).\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với x >0 và x≠1
A=(\(\dfrac{2}{\sqrt{x}-2}\)+\(\dfrac{3}{2\sqrt{x+1}}\)-\(\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\)):\(\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\) với x > 0 và x≠4
A=\(\dfrac{x+1-2\sqrt{x}}{\sqrt{x-1}}\)+\(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\) với x≥0 và x≠1
V=(\(\dfrac{1}{\sqrt{x}+2}\)+\(\dfrac{1}{\sqrt{x}-2}\))\(\dfrac{\sqrt{x}+2}{\sqrt{x}}\) với x>0 và x≠4
A=(\(\dfrac{1}{x-1}\)+\(\dfrac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\))(\(\dfrac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}\)-1)
P=\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\)+\(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}\)+\(\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
MỌI NGƯỜI GIÚP ĐỠ MÌNH RÚT GỌN MẤY BIỂU THỨC NÀY VỚI Ạ . EM XIN CẢM ƠN
a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: \(=\dfrac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)
\(=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
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\(\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0\)
a x^2-x0 b x^2-2x0 c (x+1)(x+2)(2-x)(x+2)
d dfrac{1}{x-2}+3dfrac{3-x}{x-2} đ dfrac{8-x}{x-7}-8dfrac{1}{x-7}
e dfrac{x}{2left(x-3right)}+dfrac{x}{2left(x+1right)}dfrac{2x}{left(x+1right)left(x-3right)}
f 5+dfrac{76}{x^2-16}dfrac{2x-1}{x+4}-dfrac{3x-1}{4-x}
g dfrac{90}{x}-dfrac{36}{x-6}2 h dfrac{1}{x}+dfrac{1}{x+10}dfrac{1}{12} i dfrac{x+3}{x-3}-dfrac{1}{...
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a \(x^2-x=0\) b \(x^2-2x=0\) c (x+1)(x+2)=(2-x)(x+2)
d \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\) đ \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
e \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
f \(5+\dfrac{76}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)
g \(\dfrac{90}{x}-\dfrac{36}{x-6}=2\) h \(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\) i \(\dfrac{x+3}{x-3}-\dfrac{1}{x}=\dfrac{3}{x\left(x-3\right)}\)
k \(\dfrac{3}{x+2}-\dfrac{2}{x-2}+\dfrac{8}{x^2-4}=0\) l \(\dfrac{3}{x+2}-\dfrac{2}{x-3}=\dfrac{8}{\left(x-3\right)\left(x+2\right)}\)
m\(\dfrac{x}{2x+6}-\dfrac{x}{2x+2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
n \(\dfrac{x}{x+1}-\dfrac{2x-3}{1-x}=\dfrac{3x^2+5}{x^2-1}\) j \(\dfrac{5}{x+7}+\dfrac{8}{2x+14}=\dfrac{3}{2}\)
q \(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
Cần gấp ạ
câu nào cũng ghi lại đề nha
a) \(x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)\(x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
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d) \(\dfrac{1}{x-2}+3-\dfrac{3-x}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3x-6-3+x}{x-2}=0\) ( đk \(x\ne2\) )
\(\Leftrightarrow4x-8=0\Rightarrow x=2\)
đ) \(\dfrac{8-x}{x-7}-8-\dfrac{1}{x-7}=0\)
\(\Leftrightarrow\dfrac{8-x-8\left(x-7\right)-1}{x-7}=0\) (đk \(x\ne7\))
\(\Leftrightarrow8-x-8x+56-1=0\)
\(\Leftrightarrow-9x+63=0\)
\(\Leftrightarrow x=7\)
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e) \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) (MTC 2(x+1)(x-3) )
\(\Leftrightarrow\dfrac{x\left(x+1\right)+x\left(x-3\right)-2x.2}{2\left(x-3\right)\left(x+1\right)}=0\) (đk \(x\ne3vsx\ne-1\) )
\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
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giải các phương trình sau
a, 3x -(3x+2) =x+3
b, \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
c, \(\left(x^2-3^2\right)+2\left(x-3\right)=0\)
d,\(\dfrac{1}{x-1}+\dfrac{2}{1+x}-\dfrac{4x+6}{x^2-1}=0\)
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
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