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\(A=\left(\dfrac{x+y}{y}+\dfrac{2y}{x-y}\right)\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\left(\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\right)\cdot\dfrac{1-2x}{x+2}\)

\(=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{x+2}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)

\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)

\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)

\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)

14 tháng 12 2018

\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)

\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)

\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)

15 tháng 9 2023

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)

d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)

e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)

f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)

NV
23 tháng 12 2022

\(A=\dfrac{2x}{x\left(x+y\right)}+\dfrac{6x}{\left(x-y\right)\left(x+y\right)}-\dfrac{3}{x-y}\)

\(=\dfrac{2\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}+\dfrac{6x}{\left(x-y\right)\left(x+y\right)}-\dfrac{3\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}\)

\(=\dfrac{2x-2y+6x-3x-3y}{\left(x-y\right)\left(x+y\right)}=\dfrac{5\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{5}{x+y}\)

a: \(=\left(\dfrac{x}{y\left(x-y\right)}-\dfrac{2x-y}{x\left(x-y\right)}\right):\dfrac{x+y}{xy}\)

\(=\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}\cdot\dfrac{xy}{x+y}\)

\(=\dfrac{\left(x-y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x-y}{x+y}\)

b: \(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\)

\(=\dfrac{4xy+4y^2}{2\left(x+y\right)}\cdot\dfrac{1}{2y}=\dfrac{4y\left(x+y\right)}{4y\left(x+y\right)}=1\)

19 tháng 11 2021

\(a,ĐK:x\ne0;x\ne1;x\ne\pm2\\ b,A=\left[\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{\left(2-x\right)\left(x+2\right)}\right]\cdot\dfrac{x\left(2-x\right)}{x-1}\\ A=\dfrac{x^2+4x+4-x^2+4x-4+4x^2}{\left(2-x\right)\left(x+2\right)}\cdot\dfrac{x\left(2-x\right)}{x-1}\\ A=\dfrac{4x\left(x+1\right)\cdot x}{\left(x+2\right)\left(x-1\right)}=\dfrac{4x^2}{x+2}\)