Violympic toán 8

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

24 tháng 7 2020 lúc 9:40

( ĐKXĐ tự giải hen )

ĐKXĐ : $\left\{{}\begin{matrix}x\ne\pm3\\x\ne0\end{matrix}\right.$

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

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

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

$=\left(\frac{-3x-9}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{2x+4}{x\left(x-3\right)}\right)$$=\left(\frac{-3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{2x+4}{x\left(x-3\right)}\right)$

$=\frac{-3}{\left(x-3\right)}:\frac{2x+4}{x\left(x-3\right)}$$=\frac{-3x\left(x-3\right)}{\left(x-3\right)\left(2x+4\right)}=\frac{-3x}{2x+4}$

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