a/Thay x = -2 vào A, ta có:
\(A=\dfrac{-2}{-2+3}=-2\)
b/\(M=A+B\)
\(=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+9}{x^2-9}\)
\(=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{2x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{3x^2+9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3x-9}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3}{x+3}\)
a.
\(x=-2\Rightarrow A=\dfrac{-2}{-2+3}=-2\)
b.
\(M=A+B=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+9}{x^2-9}\)
\(=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3x^2+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x\left(x-3\right)+2x\left(x+3\right)-\left(3x^2+9\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3}{x+3}\)
