a) Ta có: \(\frac{x+2}{2}-\frac{2x-3}{5}=\frac{10x+13}{10}\)
\(\Leftrightarrow\frac{5\left(x+2\right)}{10}-\frac{2\left(2x-3\right)}{10}-\frac{10x+13}{10}=0\)
Suy ra: \(5x+10-4x+6-10x-13=0\)
\(\Leftrightarrow-9x+3=0\)
\(\Leftrightarrow-9x=-3\)
hay \(x=\frac{1}{3}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{3}\right\}\)
b) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\frac{x-1}{x-2}-\frac{5}{x+2}=\frac{x^2}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x^2}{\left(x+2\right)\left(x-2\right)}=0\)
Suy ra: \(x^2+x-2-5x+10-x^2=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow-4x=-8\)
hay x=2(ktm)
Vậy: Tập nghiệm \(S=\varnothing\)