cong phan thuc
x+y/2(x-y)+2y2/y-x
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x+\(\dfrac{y}{2}\)+x+\(\dfrac{2}{2}\)x2+4
=2x+\(\dfrac{4+y}{2}\)+4
\(\dfrac{x}{x-y}-\dfrac{1}{x-y}-\dfrac{1-y}{y-x}=\dfrac{x}{x-y}-\dfrac{1}{x-y}+\dfrac{y-1}{x-y}=\dfrac{x-1+y-1}{x-y}=\dfrac{x+y-2}{x-y}\)
c: \(=\dfrac{8}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x^2+3\right)\left(x-1\right)}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x-1}\)
\(x^2+y^2-x^2y^2+xy-x-y\)\(=\left(xy-x\right)+\left(y^2-y\right)-\left(x^2y^2-x^2\right)\)
\(=x\left(y-1\right)+y\left(y-1\right)-x^2\left(y^2-1\right)\)\(=\left(y-1\right)\left[x+y-x^2\left(y+1\right)\right]\)
\(=\left(y-1\right)\left(x+y-x^2y-x^2\right)\)\(=\left(y-1\right)\left[x\left(1-x\right)+y\left(1-x^2\right)\right]\)
\(=\left(y-1\right)\left(1-x\right)\left[x+y\left(1+x\right)\right]=\left(y-1\right)\left(1-x\right)\left(xy+x+y\right)\)
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
\(\dfrac{x+y}{2\left(x-y\right)}+\dfrac{2y^2}{y-x}=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{2y^2}{x-y}=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{4y^2}{2\left(x-y\right)}=\dfrac{x+y-4y^2}{x-y}\)