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a: \(=x-\dfrac{3}{2}+2y\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
MTC = (x - y)(x2 + xy + y2)
\(\dfrac{1}{x-y}-\dfrac{3xy}{x^3-y^3}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{x^2+xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x^2-2xy+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
1/x-y-3xy/x^3-y^3+x-y/x^2+xy+y^2
=1/x-y+-3xy/(x-y)(x^2+xy+y^2)+x-y/x^2+xy+y^2
=x^2+xy+y^2/(x-y)(x^2+xy+y^2)+-3xy/(x-y)(x^2+xy+y^2)+x^2-2xy+y^2/(x-y)(x^2+xy+y^2)
=x^2+xy+y^2-3xy+x^2-2xy-y^2/(x-y)(x^2+xy+y^2)
=2x^2-5xy/(x-y)(x^2+xy+y^2)
Ta có:(x2-y2)\(.\dfrac{x^2+y^2}{y^4-x^2y^2}\)\(=\left(x^2-y^2\right).\dfrac{x^2+y^2}{y^2\left(y^2-x^2\right)}=-\dfrac{x^2+y^2}{y^2}\)
Ta có:\(\dfrac{4x^2-9y^2}{xy}:\left(2x-3y\right)=\dfrac{\left(2x-3y\right)\left(2x+3y\right)}{xy}.\dfrac{1}{\left(2x-3y\right)}=\dfrac{2x+3y}{xy}\)
\(a,=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{x-y}{xy\left(y-x\right)}=\dfrac{-1}{xy}\\ b,=\dfrac{x+3-x-4}{x-2}=\dfrac{-1}{x-2}\)
Ta có:
\(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\\ =\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{x^3-y^3}\\ =\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\\ =\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
\(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\) \(=\dfrac{x^2+xy+y^2}{x^3-y^3}-\dfrac{3xy}{x^3-y^3}+\dfrac{\left(x-y\right)^2}{x^3-y^3}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{x^3-y^3}\)
\(=\dfrac{2x^2+2y^2-4xy}{x^3-y^3}\)
\(=\dfrac{2x^2-2xy-2xy+2y^2}{x^3-y^3}\)
\(=\dfrac{2x\left(x-y\right)-2y\left(x-y\right)}{x^3-y^3}\)
\(=\dfrac{\left(2x-2y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x-2y}{x^2+xy+y^2}\)
\(\dfrac{xy}{x-y}-\dfrac{2x^2}{y-2x}\)
\(=\dfrac{xy}{x-y}+\dfrac{2x^2}{2x-y}\)
\(=\dfrac{xy\left(2x-y\right)+2x^2\left(x-y\right)}{\left(x-y\right)\left(2x-y\right)}\)
\(=\dfrac{2x^2y-xy^2+2x^3-2x^2y}{\left(x-y\right)\left(2x-y\right)}\)
\(=\dfrac{2x^3-xy^2}{\left(x-y\right)\left(2x-y\right)}=\dfrac{x\left(2x^2-y^2\right)}{\left(x-y\right)\left(2x-y\right)}\)
Ta có: \(\dfrac{y}{x-y}-\dfrac{x^3-xy^2}{x^2+y^2}\cdot\left(\dfrac{x}{x^2-2xy+y^2}-\dfrac{y}{x^2-y^2}\right)\)
\(=\dfrac{y}{x-y}-\dfrac{x\left(x^2-y^2\right)}{x^2+y^2}\cdot\left(\dfrac{x\left(x+y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)}-\dfrac{y\cdot\left(x-y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)}\right)\)
\(=\dfrac{y}{x-y}-\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2+y^2}\cdot\dfrac{x^2+xy-xy+y^2}{\left(x-y\right)^2\left(x+y\right)}\)
\(=\dfrac{y}{x-y}-\dfrac{x\cdot\left(x^2+y^2\right)}{\left(x^2+y^2\right)\cdot\left(x-y\right)}\)
\(=\dfrac{y}{x-y}-\dfrac{x}{x-y}\)
\(=\dfrac{y-x}{x-y}=\dfrac{-\left(x-y\right)}{x-y}=-1\)
\(\left(x-y\right)\left(y^2+xy+x^2\right)\)
\(=\left(x-y\right)\left(x^2+x\cdot y+y^2\right)\)
\(=x^3-y^3\)