BF

\(\)(\(\dfrac{x^2-3x}{x^2-9}-1\))  :  (\(\dfrac{9-x^2}{x^2+x-6}-\dfrac{x-3}{2-x}+\dfrac{x-2}{x+3}\))

rút gọn ạ !!!!

 

NT
4 tháng 7 2021 lúc 20:38

Ta có: \(\left(\dfrac{x^2-3x}{x^2-9}-1\right):\left(\dfrac{9-x^2}{x^2+x-6}-\dfrac{x-3}{2-x}+\dfrac{x-2}{x+3}\right)\)

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

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

\(=\dfrac{x-x-3}{x+3}\cdot\dfrac{x+3}{x-2}\)

\(=\dfrac{-3}{x-2}\)

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H24
4 tháng 7 2021 lúc 20:40

Điều kiện : x ≠ 2 ; x ≠ 3 ; x ≠ - 3

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

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

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

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

\(=\dfrac{-3}{x-2}\)

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