Question

[hello hello]

cho biểu thức M = \(\left(\dfrac{y}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

a. rút gọn M

b. Tìm x để M = 2

Answer 1

[Ngô Tấn Đạt]

\(M=\left(\dfrac{y}{x\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3\left(x-3\right)}{3.x\left(x+3\right)}-\dfrac{x^2}{3.x.\left(x+3\right)}\right)\\ =\left(\dfrac{y+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3x-9-x^2}{3.x.\left(x+3\right)}\right)\\ =\dfrac{y+x^2-3x}{x\left(x-3\right)\left(x+3\right)}.\dfrac{3.x.\left(x+3\right)}{3x-9-x^2}\\ =\dfrac{\left(y+x^2-3x\right).3}{\left(x-3\right)\left(3x-9-x^2\right)}\\ \)

Answer 2

[Shinichi Kudo]

a, ĐKXĐ: \(x\ne0;x\ne\pm3\)

Ta có: \(\left(\dfrac{y}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

\(=\left[\dfrac{y}{x\left(x^2-9\right)}+\dfrac{1}{x+3}\right]:\left[\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right]\)

\(=\left[\dfrac{y}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right]:\left[\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right]\)

\(=\dfrac{y+1.x.\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{\left(x-3\right)\left(x+3\right)-x.x}{3x\left(x+3\right)}\)

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

\(=\dfrac{y+x^2-3x}{x\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{x^2-9-x^2}\)

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