Câu trả lời:
1)\(\dfrac{6}{x^2-9}+\dfrac{5}{x-3}+\dfrac{1}{x+3}\)
=\(\dfrac{6}{\left(x-3\right)\left(x+3\right)}+\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{1\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{6+5\left(x+3\right)+1\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)=\(\dfrac{6x+18}{\left(x-3\right)\left(x+3\right)}\)=\(\dfrac{6\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)=\(\dfrac{6}{x-3}\)
2)\(\dfrac{3}{x-3}+\dfrac{-6x}{x^2-9}+\dfrac{x}{x+3}\)
=\(\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{-6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{3\left(x+3\right)+\left(-6\right)x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{3x+9+\left(-6\right)x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)=\(\dfrac{9+\left(-6\right)x+x^2}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{\left(-3\right)^2+2.\left(-3\right)x+x^2}{\left(x-3\right)\left(x+3\right)}\)=\(\dfrac{\left(-3+x\right)^2}{\left(x-3\right)\left(x+3\right)}\)=\(\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x+3}{x-3}\)