a) Ta có :
\(\frac{2x-5}{x+5}=3\)
\(\Leftrightarrow\)\(2x-5=3\left(x+5\right)\)
\(\Leftrightarrow\)\(2x-5=3x+15\)
\(\Leftrightarrow\)\(3x-2x=-5-15\)
\(\Leftrightarrow\)\(x=-20\)
Vậy \(x=-20\)
b) Ta có :
\(\frac{5}{3x+2}=2x-1\)
\(\Leftrightarrow\)\(5=\left(2x-1\right)\left(3x+2\right)\)
\(\Leftrightarrow\)\(5=3x\left(2x-1\right)+2\left(2x-1\right)\)
\(\Leftrightarrow\)\(5=6x^2-3x+4x-2\)
\(\Leftrightarrow\)\(6x^2+x=7\)
\(\Leftrightarrow\)\(x\left(6x+1\right)=7\)
TRƯỜNG HỢP 1 :
\(\hept{\begin{cases}x=1\\6x+1=7\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\6x=6\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\x=1\end{cases}}}\)
TRƯỜNG HỢP 2 :
\(\hept{\begin{cases}x=-1\\6x+1=-7\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\6x=-8\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\x=-\frac{4}{3}\end{cases}}}\)( LOẠI )
Vậy \(x=1\)
