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\(a,\dfrac{x}{x+3}+\dfrac{2-x}{x+3}\\ =\dfrac{x+2-x}{x+3}\\ =\dfrac{2}{x+3}\\b,\dfrac{x^2y}{x-y}-\dfrac{xy^2}{x-y}\\ =\dfrac{x^2y-xy^2}{x-y}\\ =\dfrac{xy\left(x-y\right)}{x-y}\\ =xy\\ c,\dfrac{2x}{2x-y}+\dfrac{y}{y-2x}\\=\dfrac{2x}{2x-y}-\dfrac{y}{2x-y}\\ =\dfrac{2x-y}{2x-y}\\ =1 \)
`a, x/(x+3) + (2-x)/(x+3) = (x+2-x)/(x+3) = 2/(x+3)`
`b, (x^2y)/(x-y) - (xy^2)/(x-y) = (x^2y-xy^2)/(x-y) = (xy(x-y))/(x-y)= xy`
`c, (2x)/(2x-y) - (y)/(2x-y)`
`= (2x-y)/(2x-y) = 1`
\(-x-y^2+x^2-y=-\left(x+y\right)-\left(x-y\right)\left(x+y\right)=\left(x+y\right)\left(-1-x+y\right)\)
a: \(=\dfrac{3b+4a}{6ab}\)
b: \(=\dfrac{x^2-2x+1-x^2-2x-1}{x^2-1}=\dfrac{-4x}{x^2-1}\)
c: \(=\dfrac{xz+yz-xy-xz}{xyz}=\dfrac{yz-xy}{xyz}=\dfrac{z-x}{xz}\)
d: \(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
e: \(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)
Câu 1 : x2-y2+2yz-z2=-(y2-2yz+z2-x2) Câu 2: x2-2xy+y2-xz+yz=(x2-2xy+y2)-xz+yz
=-(y-z)2 -x2 =(x-y)2-z(x-y)
=-(y-z-x)(y-z+x) =(x-y)(x-y-z)