ĐKXĐ : \(x\ne\frac{3}{2};x\ne-\frac{7}{4}\)
Có \(\frac{4x^2+3x-7}{A}=\frac{4x+7}{2x-3}\)
\(\Leftrightarrow A=\frac{\left(4x^2+3x-7\right)\left(2x-3\right)}{4x+7}\)
\(\Leftrightarrow A=\frac{\left(4x^2-4x+7x-7\right)\left(2x-3\right)}{4x+7}\)
\(\Leftrightarrow A=\frac{\left(4x+7\right)\left(x-1\right)\left(2x-3\right)}{4x+7}\)
\(\Leftrightarrow A=\left(x-1\right)\left(2x-3\right)=2x^2-5x+3\)
Ta có: \(\frac{4x^2+3x-7}{A}=\frac{4x+7}{2x-3}\)
\(\Leftrightarrow\frac{4x^2+7x-4x-7}{A}=\frac{4x+7}{2x-3}\)
\(\Leftrightarrow\frac{x\left(4x+7\right)-\left(4x+7\right)}{A}=\frac{4x+7}{2x-3}\)
\(\Leftrightarrow\frac{\left(4x+7\right)\left(x-1\right)}{A}=\frac{4x+7}{2x-3}\)
\(\Leftrightarrow A=\frac{\left(4x+7\right)\left(x-1\right)\left(2x-3\right)}{4x+7}\)
\(\Leftrightarrow A=\left(x-1\right)\left(2x-3\right)\)
\(\Leftrightarrow A=2x^2-3x-2x+3\)
hay \(A=2x^2-5x+3\)
Vậy: \(A=2x^2-5x+3\)
