Question

rút gọn biểu thức

\(\left(\dfrac{x-1}{x-3}-\dfrac{1-x}{x+3}\right).\dfrac{3}{x^2+3}\)

Answer 1

\(\left(\dfrac{x-1}{x-3}-\dfrac{1-x}{x+3}\right).\dfrac{3}{x^2+3}=\left[\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right].\dfrac{3}{x^2+3}\)

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

\(=\dfrac{2x^2-2x}{\left(x-3\right)\left(x+3\right)}.\dfrac{3}{\left(x^2+3\right)}=\dfrac{6x\left(x-1\right)}{\left(x-3\right)\left(x+3\right)\left(x^2+3\right)}\)