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先化简再求值 [(1/(x-y)+1/(x+y)]-【2y/(x的平方-2xy+y的平方)】 其中x=1+√2,y=1-√2
人气:179 ℃ 时间:2019-08-20 08:48:03
解答
x=1+√2,y=1-√2
x+y=2 x-y=2√2
[(1/(x-y)+1/(x+y)]-【2y/(x的平方-2xy+y的平方)】
=[(x+y+x-y/(x-y)(x+y)]-【2y/(x-y)^2】
=[2(1+√2)/2*2√2]-【2(1-√2)/(2√2)^2】
=[(1+√2)/2√2]-【2(1-√2)/8】
=[(2+√2)/4]-【(1-√2)/4】
=(1+2√2)/4难道不需要化到最简再带入么x=1+√2,y=1-√2x+y=2x-y=2√2[(1/(x-y)+1/(x+y)]-【2y/(x的平方-2xy+y的平方)】 =[(x+y+x-y/(x-y)(x+y)]-【2y/(x-y)^2】=[(2x/(x-y)(x+y)]-【2y/(x-y)^2】=(2x(x-y)-2y(x+y))/(x+y)(x-y)^2=(2x^2-2xy-2xy-2y^2)/(x+y)(x-y)^2=2(x^2-2xy-y^2)/(x+y)(x-y)^2应该化成最简,当时看化简意义不大,才直接代入x、y的值
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