\(\frac{x+1}{2x-3}=\frac{x+3}{x-3}\)
\(\Leftrightarrow\left(x+1\right).\left(x-3\right)=\left(2x-3\right).\left(x+3\right)\)
\(\Leftrightarrow x^2-3x+x-3=2x^2+6x-3x-9\)
\(\Leftrightarrow x^2-3x+x-3-2x^2-6x+3x+9=0\)
\(\Leftrightarrow-x^2-5x+6=0\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow x^2-x+6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+6=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-6\end{array}\right.\)
\(\frac{x+1}{2x-3}=\frac{x+3}{x-3}\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=\left(x+3\right)\left(2x-3\right)\)
\(\Leftrightarrow x^2-3x+x-3=2x^2-3x+6x-9\)
\(\Leftrightarrow\left(x^2-2x^2\right)+\left(x-6x\right)+\left(3x-3x\right)=\left(-9+3\right)\)
\(\Leftrightarrow-x^2-5x=-6\)
\(\Leftrightarrow x\left(x+5\right)=6\)
Giải với các giá trị \(x\inƯ_6\)
Ta được x=1 ; x= - 6
