
数学问题...分式
(x/x+y+2y/x+y)*xy/x+2y÷(1/x+1/y)(1/a+1/b)²÷(1/a²-1/b²)...
( x/x+y + 2y/x+y)* xy/x+2y ÷(1/x + 1/y)
( 1/a + 1/b)²÷(1/a² -1/b²) 展开
( 1/a + 1/b)²÷(1/a² -1/b²) 展开
2个回答
展开全部
( 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)÷(x+y)/xy
=x^2y^2/(x+y)^2
( 1/a + 1/b)^2÷(1/a^2 -1/b^2)
=[(a+b)/ab]^2÷(b^2-a^2)/a^2b^2
=(a+b)^2/a^2b^2×a^2b^2/(b-a)(b+a)
=(a+b)/(b-a)
=(x+2y)/(x+y) *xy/(x+2y)÷(x+y)/xy
=xy/(x+y)÷(x+y)/xy
=x^2y^2/(x+y)^2
( 1/a + 1/b)^2÷(1/a^2 -1/b^2)
=[(a+b)/ab]^2÷(b^2-a^2)/a^2b^2
=(a+b)^2/a^2b^2×a^2b^2/(b-a)(b+a)
=(a+b)/(b-a)
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