\(\frac{x+1}{2x-3}=\frac{x+3}{x-3}\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=\left(2x-3\right)\left(x+3\right)\)
\(\Leftrightarrow x^2-3x+x-3=2x^2+6x-3x-9\)
\(\Leftrightarrow x^2-3x+x-3-2x^2-6x+3x+9=0\)
\(\Leftrightarrow-x^2-5x+6=0\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow x^2-x+6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+6=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-6\end{array}\right.\)
Điều kiện : \(x\ne3;x\ne\frac{3}{2}\)
\(\frac{x+1}{2x-3}=\frac{x+3}{x-3}\Leftrightarrow\left(x+1\right)\left(x-3\right)=\left(2x-3\right)\left(x+3\right)\)
\(\Leftrightarrow x^2-2x-3=2x^2+3x-9\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\Leftrightarrow\left[\begin{array}{nghiempt}x=-6\\x=1\end{array}\right.\) (thỏa mãn)
\(\Rightarrow\left(x+1\right)\left(x-3\right)=\left(x+3\right)\left(2x-3\right)\)
=> \(x^2-3x+x-3=2x^2-3x+6x-9\)
=> \(-3+9=2x^2-3x+6x-x^2+3x-x\)
=> \(x^2+5x=6\)
=> \(x\left(x+5\right)=6\)
THử với các ước của 6 ta được x=1
Vậy x=1
\(\frac{x+1}{2x-3}=\frac{x+3}{x-3}\)
=>\(\left(x+1\right)\left(x-3\right)=\left(x+3\right)\left(2x-3\right)\)
=>\(x^2+x-3x-3=2x^2+6x-3x-9\)
=> \(\left(x^2+x-3x-3\right)-\left(2x^2+6x-3x-9\right)=0\)
=>\(x^2+x-3x-3-2x^2-6x+3x+9=0\)
=>\(\left(x^2-2x^2\right)+\left(x-3x-6x+3x\right)-\left(3-9\right)=0\)
=>\(\left(-x\right)^2-5x+6=0\)
hay \(x^2-5x+6=0\)
Ta có: \(x^2-5x+6=0\)
=>\(x^2-\left(2x+3x\right)+6=0\)
=>\(x^2-2x-3x+6=0\)
=> \(\left(x^2-2x\right)-\left(3x-6\right)=0\)
=>\(x.\left(x-2\right)-3.\left(x-2\right)=0\)
=>\(\left(x-3\right).\left(x-2\right)=0\)
=> x-3=0 hoặc x-2=0
=> x=3 hoặc x=2
