a ) \(\dfrac{x-y}{x^3+y^3}.Q=\dfrac{x^2-2xy+y^2}{x^2-xy+y^2}\)
\(\Leftrightarrow Q=\dfrac{x^2-2xy+y^2}{x^2-xy+y^2}:\dfrac{x-y}{x^3+y^3}\)
\(\Leftrightarrow Q=\dfrac{\left(x-y\right)^2}{x^2-xy+y^2}\cdot\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x-y}\)
\(\Rightarrow Q=\left(x-y\right)\left(x+y\right)=x^2-y^2\)
Vậy \(Q=x^2-y^2\)
b ) \(\dfrac{x+y}{x^3-y^3}.Q=\dfrac{3x^2+3xy}{x^2+xy+y^2}\)
\(\Leftrightarrow Q=\dfrac{3x^2+3xy}{x^2+xy+y^2}:\dfrac{x+y}{x^3-y^3}\)
\(\Leftrightarrow Q=\dfrac{3x\left(x+y\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{x+y}\)
\(\Leftrightarrow Q=3x\left(x-y\right)=3x^2-3xy\)
Vậy \(Q=3x^2-3xy\)
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