a) Ta có: \(B=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
\(=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)
b)
ĐKXĐ: \(x\notin\left\{-3;2\right\}\)
Để \(B=\frac{3}{2}\) thì \(\frac{x-4}{x-2}=\frac{3}{2}\)
\(\Leftrightarrow2\left(x-4\right)=3\left(x-2\right)\)
\(\Leftrightarrow2x-8=3x-6\)
\(\Leftrightarrow2x-8-3x+6=0\)
\(\Leftrightarrow-x-2=0\)
\(\Leftrightarrow-x=2\)
hay x=-2(tm)
Vậy: Khi \(B=\frac{3}{2}\) thì x=-2
