\(\frac{y+5}{y^2-5y}-\frac{y-5}{2y^2+10y}=\frac{y+25}{2y^2-50}\) \(ĐKXĐ:y\ne0;y\ne5\)
\(\Leftrightarrow\frac{y+5}{y\left(y-5\right)}-\frac{y-5}{2y\left(y+5\right)}-\frac{y+25}{2\left(y^2-25\right)}=0\)
\(\Leftrightarrow\frac{2\left(y+5\right)^2-\left(y-5\right)^2}{2y\left(y-5\right)\left(y+5\right)}-\frac{y\left(y+25\right)}{2y\left(y-5\right)\left(y+5\right)}=0\)
\(\Leftrightarrow\frac{2\left(y^2+10y+25\right)-y^2+10y-25-y^2-25y}{2y\left(y-5\right)\left(y+5\right)}=0\)
\(\Leftrightarrow2y^2+20y+50-y^2+10y-25-y^2-25y=0\)
\(\Leftrightarrow5y+25=0\)
\(\Leftrightarrow5\left(y+5\right)=0\)
\(\Leftrightarrow y=-5\left(tm\right)\)
\(\)Vậy phương trình có nghiệm y=-5
Sai rồi nha bae , ĐKXĐ là \(\hept{\begin{cases}y\ne0\\y\ne\pm5\end{cases}}\)nha nên dẫn đến đáp án sai luôn
