Question

Giúp nhanh nhé

Answer 1

a: \(2y^2-3y=2\)

=>\(2y^2-3y-2=0\)

=>\(2y^2-4y+y-2=0\)

=>2y(y-2)+(y-2)=0

=>(y-2)(2y+1)=0

=>\(\left[{}\begin{matrix}y=2\\y=-\dfrac{1}{2}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}x+2y=3\\2x+y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+4y=6\\2x+y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3y=3\\2x+y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=1\\2x=3-y=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)