\(\left(2x-3\right)^2=5\Rightarrow x\ne0\)
\(4x^2-12x+9=5\)
\(\Leftrightarrow x^2+1=3x\)\(\Leftrightarrow x+\frac{1}{x}=3\)
\(\Rightarrow\left(x+\frac{1}{x}\right)^2=9\Rightarrow x^2+2+\frac{1}{x^2}=9\)
\(\Rightarrow x^2+\frac{1}{x^2}=7\)
Ta có:
\(A=\frac{3}{x^2+\frac{1}{x^2}-9}=\frac{3}{7-9}=-\frac{3}{2}\)