mn giúp e vs ạk ($\frac{2x+1}{x^{2}-5x}$ -$\frac{2x}{x^{2}+5x}$) .$\frac{x^{3}-25x}{21x-2}$

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1. Đáp án:

   $\dfrac{21x +5}{21x -2}$

   Giải thích các bước giải:

   $\left(\dfrac{2x+1}{x^2 – 5x} -\dfrac{2x}{x^2 + 5x}\right)\cdot \dfrac{x^3 – 25x}{21x -2}\qquad (*)$

   $ĐKXĐ:\begin{cases}x^2-5x \ne 0\\x^2 +5x \ne 0\\21x -2\ne 0\end{cases}\longrightarrow \begin{cases}x \ne 0\\x \ne \pm 5\\x \ne \dfrac{2}{21}\end{cases}$

   $(*)=\left[\dfrac{(2x+1)(x+5)}{x(x-5)(x+5)}-\dfrac{2x(x-5)}{x(x-5)(x+5)}\right]\cdot\dfrac{x(x-5)(x+5)}{21x -2}$

   $= \dfrac{2x^2 + 11x + 5 – (2x^2 – 10x)}{x(x-5)(x+5)}\cdot\dfrac{x(x-5)(x+5)}{21x -2}$

   $= \dfrac{21x +5}{21x -2}$

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2. ($\frac{2x+1}{x^{2}-5x}$-$\frac{2x}{x^{2}+5x}$).$\frac{x^{3}-25x}{21x-2}$

   =($\frac{2x+1}{x(x-5)}$-$\frac{2x}{x(x+5)}$). $\frac{x(x^{2}-25)}{21x-2}$

   =$\frac{(2x+1)(x+5)-2x(x-5)}{x(x-5)(x+5)}$.$\frac{x(x-5)(x+5)}{21x-2}$

   =$\frac{2x^{2}+10x+x+5-2x^{2}+10x}{x(x+5)(x-5)}$.$\frac{x(x+5)(x-5)}{21x-2}$

   =$\frac{21x+5}{x(x+5)(x-5)}$.$\frac{x(x+5)(x-5)}{21x-2}$

   =$\frac{21x+5}{21x-2}$ 

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