Bài 4:
a) Ta có; \(x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Thay `x=0` vào A ta có:
\(A=\dfrac{0-3}{0+1}=-3\)
b) \(B=\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\)
\(=\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}\)
\(=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{6x}{\left(x+3\right)\left(x-3\right)}+\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3x+9+6x+x^2-3x}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2+6x+9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x+3}{x-3}\)
