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Huỳnh Kiệt

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

Giải cụ thể, từng bước dùm nha. Cảm ơn nhiều

Nguyễn Nam
8 tháng 12 2017 lúc 17:16

\(\left(\dfrac{9}{x^2-9}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\) ( sửa đề \(x^3-9\) thành \(x^2-9\) )

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

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

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

\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{3\left(x-3\right)}{3x\left(x+3\right)}-\dfrac{x.x}{3x\left(x+3\right)}\right)\)

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

\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}\)

\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)

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

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

Huỳnh Kiệt
8 tháng 12 2017 lúc 18:03

ngay chỗ x^3-9 phải là x^3-9x nha

Huỳnh Kiệt
8 tháng 12 2017 lúc 18:06

ai giải giúp mình nha

Nguyễn Nam
8 tháng 12 2017 lúc 19:44

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

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

\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

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

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

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

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

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

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

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

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

\(=\dfrac{-3}{x-3}\)


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