(x/2y)^2*y/2x-x/y^2/2y^2/x(x/x+y+2y/x+y)*xy/x+2y/(1/x+1/y)(1/a+1/b)^2/(1/a^2-1/b^2)
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(x/2y)^2*y/2x-x/y^2/2y^2/x(x/x+y+2y/x+y)*xy/x+2y/(1/x+1/y)(1/a+1/b)^2/(1/a^2-1/b^2)(x/2y)^2*y/2x-x/y
(x/2y)^2*y/2x-x/y^2/2y^2/x(x/x+y+2y/x+y)*xy/x+2y/(1/x+1/y)(1/a+1/b)^2/(1/a^2-1/b^2)
(x/2y)^2*y/2x-x/y^2/2y^2/x
(x/x+y+2y/x+y)*xy/x+2y/(1/x+1/y)
(1/a+1/b)^2/(1/a^2-1/b^2)
(x/2y)^2*y/2x-x/y^2/2y^2/x(x/x+y+2y/x+y)*xy/x+2y/(1/x+1/y)(1/a+1/b)^2/(1/a^2-1/b^2)
(x/2y)^2*y/2x-x/y^2/(2y^2/x)
=x^2/4y^2*y/2x-x/y^2*(x/2y^2)
=x/8y-x^2/2y^4
(x/x+y+2y/x+y)*xy/x+2y/(1/x+1/y)
=(x+2y)/(x+y)*xy/(x+2y)/[(x+y)/xy]
=xy/(x+y)*xy/(x+y)
=x^2y^2/(x+y)^2
(1/a+1/b)^2/(1/a^2-1/b^2)
=(1/a+1/b)^2/[(1/a+1/b)(1/a-1/b)]
=(1/a+1/b)/(1/a-1/b)
=(a+b)/ab/[(b-a)/ab]
=(a+b)/(b-a)
第一和第二题有点看不清楚分子与分母.
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化简:[(x+y)(x-y)(x+y)^2-2y(x-2y)](-2y) 谢谢!
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