Kết quả rút gọn biểu thức \(Q=\left(\dfrac{x^2+3x}{x^3+3x^2+9x+27}+\dfrac{3}{x^2+9}\right):\left(\dfrac{1}{x-3}-\dfrac{6x}{x^3-3x^2+9x-27}\right)\) là
\(\dfrac{1}{x^2+9}.\)\(\dfrac{x-3}{x+3}.\)\(\dfrac{1}{x+3}.\)\(\dfrac{x+3}{x-3}.\)Hướng dẫn giải:ĐKXĐ: \(x\ne\pm3.\)
\(Q=\left(\dfrac{x^2+3x}{x^3+3x^2+9x+27}+\dfrac{3}{x^2+9}\right):\left(\dfrac{1}{x-3}-\dfrac{6x}{x^3-3x^2+9x-27}\right)\)
\(=\left(\dfrac{x\left(x+3\right)}{\left(x^2+9\right)\left(x+3\right)}+\dfrac{3}{x^2+9}\right):\left(\dfrac{1}{x-3}-\dfrac{6x}{\left(x^2+9\right)\left(x-3\right)}\right)\)
\(=\left(\dfrac{x}{x^2+9}+\dfrac{3}{x^2+9}\right):\dfrac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)
\(=\dfrac{x+3}{x^2+9}.\dfrac{\left(x-3\right)\left(x^2+9\right)}{\left(x-3\right)^2}=\dfrac{x+3}{x-3}.\)
