Nếu \(\left[{}\begin{matrix}a=0\Rightarrow b=0\Rightarrow b=2a\\b=0\Rightarrow a=0\Rightarrow b=2a\end{matrix}\right.\) trái với giả thiết \(\Rightarrow ab\ne0\)
\(2a^2+11ab-3b^2=0\Rightarrow2\left(\frac{a}{b}\right)^2+11\left(\frac{a}{b}\right)-3=0\)
Đặt \(\frac{a}{b}=x\ne0;\pm\frac{1}{2}\Rightarrow2x^2+11x-3=0\Rightarrow11x=3-2x^2\)
\(T=\frac{a-2b}{2a-b}+\frac{2a-3b}{2a+b}=\frac{\frac{a}{b}-2}{\frac{2a}{b}-1}+\frac{\frac{2a}{b}-3}{\frac{2a}{b}+1}=\frac{x-2}{2x-1}+\frac{2x-3}{2x+1}\)
\(T=\frac{\left(x-2\right)\left(2x+1\right)+\left(2x-3\right)\left(2x-1\right)}{4x^2-1}=\frac{6x^2-11x+1}{4x^2-1}=\frac{6x^2-\left(3-2x^2\right)+1}{4x^2-1}\)
\(T=\frac{8x^2-2}{4x^2-1}=2\)
