Question

Câu 14. Rút gọn biểu thức: \( A = \frac{3x+15}{x^2-9} + \frac{1}{x+3} - \frac{2}{x-3} \) (với \( x \neq \pm 3 \)).

Answer 1

\(A=\dfrac{3x+15}{x^2-9}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\)

\(=\dfrac{3x+15}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\)

\(=\dfrac{3x+15+x-3-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{4x+12-2x-6}{\left(x-3\right)\cdot\left(x+3\right)}=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\)

Answer 2

\(A=\dfrac{3x+15}{x^2-9}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\\ =\dfrac{3x+15}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\\ =\dfrac{3x+15+x-3-2\cdot\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x+15+x-3-2x-6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{2\cdot\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\)