`((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)`
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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