a: ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};\dfrac{1}{5};-\dfrac{1}{5}\right\}\)
b: Sửa đề: của biểu thức A tại x=3/5
\(A=\dfrac{5x+1}{2x-3}\cdot\dfrac{x+2}{25x^2-1}:\dfrac{2x-3}{5x+1}\)
\(=\dfrac{5x+1}{2x-3}\cdot\dfrac{x+2}{\left(5x-1\right)\left(5x+1\right)}\cdot\dfrac{5x+1}{2x-3}\)
\(=\dfrac{\left(x+2\right)}{\left(2x-3\right)^2}\)
Khi x=3/5 thì \(A=\dfrac{\left(\dfrac{3}{5}+2\right)}{\left(2\cdot\dfrac{3}{5}-3\right)^2}=\dfrac{13}{5}:\left(\dfrac{6}{5}-3\right)^2\)
\(=\dfrac{13}{5}:\left(-\dfrac{9}{5}\right)^2\)
\(=\dfrac{13}{5}:\dfrac{81}{25}=\dfrac{13}{5}\cdot\dfrac{25}{81}=\dfrac{65}{81}\)