Câu a :
\(\left(3x^2-3y^2\right)+\left(4x-4y\right)=0\)
\(\Leftrightarrow3\left(x^2-y^2\right)+4\left(x-y\right)=0\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)+4\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[3\left(x+y\right)+4\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\3\left(x+y\right)+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=-\dfrac{4}{3}\Rightarrow x=-\dfrac{4}{3}-y\end{matrix}\right.\)
Vậy \(x=y\) hoặc \(x=-\dfrac{4}{3}-y\)
Câu b :
\(\left(12x^2-3xy\right)+\left(8x-2y\right)=0\)
\(\Leftrightarrow3x\left(4x-y\right)+2\left(4x-y\right)=0\)
\(\Leftrightarrow\left(4x-y\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-y=0\\3x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{y}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{y}{4}\) hoặc \(x=-\dfrac{2}{3}\)
