2x-1 /1-x+1=1/1-x
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M
1
21 tháng 5 2022
a: \(=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{2x^2-x^3}{x^2-3x}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{-4x^2-8x}{x+2}\cdot\dfrac{-x}{x-3}\)
\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)
b: \(=\dfrac{2x-1}{2x+1}:\left(2x-1+\dfrac{2-4x}{2x+1}\right)\)
\(=\dfrac{2x-1}{2x+1}:\dfrac{4x^2-1+2-4x}{2x+1}\)
\(=\dfrac{2x-1}{4x^2-4x+1}=\dfrac{1}{2x-1}\)
c: \(=\left(\dfrac{1}{1-x}-1\right):\left(x+1-\dfrac{2x-1}{x-1}\right)\)
\(=\dfrac{1-1+x}{1-x}:\dfrac{x^2-1-2x+1}{x-1}\)
\(=\dfrac{-x}{x-1}\cdot\dfrac{x-1}{x\left(x-2\right)}=\dfrac{-1}{x-2}\)
TQ
27 tháng 1 2019
a)⇔(2x+1)(2x+1)/(2x-1)(2x+1)-(2x-1)(2x-1)/(2x-1)(2x+1)=8/(2x-1)(2x+1)
⇔(2x+1)^2-(2x-1)^2=8
⇔[(2x+1)-(2x-1)][(2x+1)(2x-1)]=8
⇔2.4x=8
⇔x=1.S={1}
20 tháng 1 2023
a: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)
=>(2x+1)^2-(2x-1)^2=8
=>4x^2+4x+1-4x^2+4x-1=8
=>8x=8
=>x=1
b: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x\cdot\left(1-\dfrac{x-1}{x+1}\right)\)
=>\(\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}=3x\cdot\dfrac{x+1-x+1}{x+1}\)
=>\(\dfrac{4x}{\left(x-1\right)\left(x+1\right)}=3x\cdot\dfrac{2}{x+1}\)
=>4x=6x(x-1)
=>6x^2-6x-4x=0
=>6x^2-10x=0
=>2x(3x-5)=0
=>x=0 hoặc x=5/3