Question

giải phương trinhf
y/2(y-3) = 2y/(y+1)(y-3) - y / 2y+ 2

Answer 1

\(\dfrac{y}{2\left(y-3\right)}=\dfrac{2y}{\left(y+1\right)\left(y-3\right)}-\dfrac{y}{2y+2}\)

\(DKXD:y\ne-1;y\ne3\)

<=>\(\dfrac{y\left(y+1\right)}{2\left(y-3\right)\left(y+1\right)}=\dfrac{4y}{2\left(y-3\right)\left(y+1\right)}-\dfrac{y\left(y-3\right)}{2\left(y+1\right)\left(y-3\right)}\)

=>y²+y=4y-y²+3y

<=>2y²-6y=0

<=>2y(y-3)=0

\(\left[{}\begin{matrix}y=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=0\left(TM\right)\\y=3\left(KTM\right)\end{matrix}\right.\)

vay..........