RÚt gọn:$\left(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\dfrac{1}{x}-\dfrac{2}{x^2}\right)$

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1. `((x^2-2x)/(2x^2+8) – (2x^2)/(8-4x+2x^2-x^3))(1 – 1/x – 2/x^2)`

   `= x/(2(x-2))(-1/x – 2/x^2 +1)`

   `= x(1 – 1/x – 2/x^2) : 2(x-2)`

   `= (x^2-x-2)/x^2x : 2(x-2)`

   `= (x^2-x-2)/x : 2(x-2)`

   `= (x^2-x-2)/(2x(x-2))`

   `= ((x+1)(x-2))/(2x(x-2))`

   `= (x+1)/(2x)`

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2. Đáp án:

    

   Giải thích các bước giải:

      ($\frac{x²-2x}{2x²+8}$-$\frac{2x²}{8-4x+2x²-x³}$)(1-$\frac{1}{x}$-$\frac{2}{x²}$)

   = ($\frac{x(x-2)}{2(x²+4)}$+$\frac{2x²}{x³-2x²+4x-8}$)($\frac{x²-x-2}{x²}$)

   = ($\frac{x(x-2)²}{2(x²+4)(x-2)}$+$\frac{4x²}{2(x²+4)(x-2)}$)($\frac{(x-2)(x+1)}{x²}$)

   = $\frac{x³-4x²+4x+4x²}{2(x²+4)(x-2)}$ . $\frac{(x-2)(x+1)}{x²}$

   = $\frac{x(x²+4)}{2(x²+4)(x-2)}$ . $\frac{(x-2)(x+1)}{x²}$

   = $\frac{x+1}{2x}$

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