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PN
2
10 tháng 6 2018
\(2\frac{2}{x-1}=1+\frac{2x}{x+2}\) \(\left(x\ne1;x\ne-2\right)\)
\(\Rightarrow\frac{2\left(x-1\right)+2}{x-1}=\frac{\left(x+2\right)+2x}{x+2}\)\(\Rightarrow2x^2+4x=3x^2+2x-3x+2\)
\(\Rightarrow\frac{2x-2+2}{x-1}=\frac{x+2+2x}{x+2}\)
\(\Rightarrow\frac{2x}{x-1}=\frac{3x+2}{x+2}\)
\(\Rightarrow2x\left(x+2\right)=\left(x-1\right)\left(3x+2\right)\)
\(\Rightarrow2x^2+4x=x\left(3x+2\right)-1\left(3x+2\right)\)
\(\Rightarrow2x^2+4x=x\left(3x+2\right)-1\left(3x+2\right)\)
HH
10 tháng 6 2018
\(2\frac{2}{x-1}=1+\frac{2x}{x+2}\) ĐKXĐ: \(\hept{\begin{cases}x\ne1\\x\ne-2\end{cases}}\)
=> \(\frac{2\left(x-1\right)+2}{x-1}=\frac{x+2+2x}{x+2}\)
=> \(\frac{2\left(x-1+1\right)}{x-1}=\frac{x+2\left(x+1\right)}{x+2}\)
=> \(\frac{2x}{x-1}=\frac{x+2\left(x+1\right)}{x+2}\)
=> \(2x\left(x+2\right)=x+2\left(x+1\right)\left(x-1\right)\)
=> \(2x^2+4x=x+2\left(x^2-1\right)\)
=> \(2x^2+4x=x+2x^2-2\)
=> \(2x^2+4x-x-2x^2+2=0\)
=> \(3x+2=0\)
=> \(3x=-2\)
=> \(x=-\frac{2}{3}\)
5 tháng 4 2022
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
8 tháng 6 2023
=>-2x+7-3x^2-x=0
=>-3x^2-3x+7=0
=>\(x=\dfrac{-3\pm\sqrt{93}}{6}\)
MP
8 tháng 6 2023
\(c.-\left(2x-7\right)-x\left(3x+1\right)=0\)
\(\Leftrightarrow-2x+7-3x^2-x=0\)
\(\Leftrightarrow-3x^2-3x+7=0\)
Vậy pt này vô n0
2 tháng 7 2019
\(a,\frac{7}{x+2}=\frac{3}{x-5}\)
\(\Rightarrow7\left(x-5\right)=3\left(x+2\right)\)
\(\Rightarrow7x-35=3x+6\)
\(\Rightarrow7x-3x=6+35\)
\(\Rightarrow4x=41\)
\(\Rightarrow x=\frac{41}{4}\)
\(b,\frac{2x+5}{2x}-\frac{x}{x+5}=0\)
\(\Rightarrow\frac{2x+5}{2x}=\frac{x}{x+5}\)
\(\Rightarrow\left(2x+5\right)\left(x+5\right)=2x\cdot x\)
\(\Rightarrow2x^2+10x+5x+25=2x^2\)
\(\Rightarrow2x^2+15x+25-2x^2=0\)
\(\Rightarrow15x+25=0\)
\(\Rightarrow15x=-25\)
\(\Rightarrow x=\frac{-5}{3}\)
\(c,\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\)
\(\Rightarrow\frac{12x+1}{11x-4}=\frac{20x+17}{18}-\frac{10x-4}{9}\)
\(\Rightarrow\frac{12x+1}{11x-4}=\frac{25}{18}\)
\(\Rightarrow\left(12x+1\right)\cdot18=25\cdot\left(11x-4\right)\)
\(\Rightarrow216x+18=275x-100\)
\(\Rightarrow216x-275x=-100-18\)
\(\Rightarrow-59x=-118\)
\(\Rightarrow x=2\)
NT
8 tháng 7 2019
A=(\(\frac{x^3-1}{x\left(x-1\right)}\)-\(\frac{x^3-1}{x\left(x+1\right)}\)) : \(\frac{2\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\)ĐKXĐ: x\(\ne\) -1, 1
A=\(\frac{1}{x\left(x+1\right)}\)x \(\frac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-1\right)}\)
A=\(\frac{1}{2x^2-2x}\)
B=\(\frac{x+1}{x-2}\)-\(\frac{2x}{x+2}\)-\(\frac{2+5x}{x^2-4}\)ĐKXĐ : x\(\ne\)2, -2
B=\(\frac{x+1}{x-2_{ }}\)-\(\frac{2x}{x+2}\)-\(\frac{2+5x}{\left(x-2\right)\left(x+2\right)}\)
B=\(\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}\)-\(\frac{2x^2-4x}{\left(x-2\right)\left(x+2\right)}\)-\(\frac{2+5x}{\left(x-2\right)\left(x+2\right)}\)
B=\(\frac{-x^2+2x}{\left(x-2\right)\left(x+2\right)}\)
B=\(\frac{-x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
B=\(\frac{-x}{x+2}\)
\(|\frac{1}{2}x+1|-4=0\)
\(\Rightarrow|\frac{1}{2}x+1|=4\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x+1=4\\\frac{1}{2}x+1=-4\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=-5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=6\\x=-10\end{cases}}\)
Vậy x = 6 hoặc x = -10
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