a) ĐKXĐ: \(x\notin\left\{-1;2;3\right\}\)
Ta có: \(E=\left(1-\frac{x}{x+1}\right):\left(\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\right)\)
\(=\left(\frac{x+1}{x+1}-\frac{x}{x+1}\right):\left(\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-4}{\left(x-3\right)\left(x-2\right)}+\frac{x+2}{\left(x-3\right)\left(x-2\right)}\right)\)
\(=\frac{1}{x+1}:\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\frac{1}{x+1}:\frac{x-3}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{1}{x+1}:\frac{1}{x-2}\)
\(=\frac{1}{x+1}\cdot\frac{x-2}{1}=\frac{x-2}{x+1}\)
b) Thay \(x=-1\frac{1}{2}=-\frac{3}{2}\) vào biểu thức \(E=\frac{x-2}{x+1}\), ta được:
\(\left(-\frac{3}{2}-2\right):\left(\frac{-3}{2}+1\right)\)
\(=\left(-\frac{3}{2}-\frac{4}{2}\right):\left(-\frac{3}{2}+\frac{2}{2}\right)\)
\(=\frac{-7}{2}:\frac{-1}{2}\)
\(=\frac{-7}{2}\cdot\frac{2}{-1}=\frac{-7}{-1}=7\)
Vậy: 7 là giá trị của biểu thức \(E=\frac{x-2}{x+1}\) tại \(x=-1\frac{1}{2}=-\frac{3}{2}\)
