Question

运用分式的加减（二）计算 (x/x+y+2y/x+y)*xy/x+2y÷(1/x+1/y)

（1）(x/x+y+2y/x+y)*xy/x+2y÷(1/x+1/y) （2） (1/a+1/b）²÷（1/a²-1/b²) （3） (3x²/4y)² * 2y/3x+x²/2y²÷2y²/x （4） (a+b/a-b)² * 2a-2b/3a+3b-a²/a²-b²÷a/b

Answer 1

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（1）原式=（x+2y)/(x+y)*xy/(x+2y)÷(x+y)/xy=xy/(x+y)*xy/(x+y)=x^2y^2/(x^2+2xy+Y^2
(2)原式=[（a+b)/ab]^2÷(b^2-a^2)/a^2b^2=(a+b)^2/a^2b^2*a^2b^2/[(a+b)(b-a)]
=(a+b)/(b-a)
(3)原式=9x^4/16y^2*2y/3x+x^2/2y^2*x/2y^2=3x^3/8y+x^3/4y^4=(3x^3y^3+2x^3)/8y^4
(4)原式=（a+b)^2/(a-b)^2*2(a-b)/3(a+b)-a^2/[(a+b)(a-b)]*b/a
=2(a+b)/3(a-b)-ab/([(a+b)(a-b)]
=(2a^2+2ab+b^2-3a^2b)/(3a^2-3b^2)