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化简(2x^2+2x/x^2-1-x^2-x/x^2-2x+1)÷x/x+1,当x=1+√2,原代数式的值能等于-1吗
人气:199 ℃ 时间:2019-09-27 12:55:49
解答
(2x^2+2x/x^2-1-x^2-x/x^2-2x+1)÷x/x+1当x=1+√2 原式=[2x(x+1)/(X+1)(X-1)-(x-1)x/(x-1)²]乘以(x+1)/x={[2X/(X-1)]- [x/(x-1)]}×(x+1)/x= x/(x-1) × (x+1)/x =(x+1)/(x-1) 带入1+√2得出原式=√2+1...
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