a/ \(\dfrac{2}{3}a^2b\left(3ab-a^2+b\right)=6a^3b^2-\dfrac{2a^2}{3}+\dfrac{2a^2b^2}{3}\)
b/ \(\left(x+y\right)y+x\left(x-y\right)=xy+y^2+x^2-xy=x^2+y^2\)
c/ \(2x\left(x-1\right)+x\left(5-2x\right)=15\)
\(\Leftrightarrow2x^2-2x+5x-2x^2=15\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
Vậy..