[1+1/(x-1)]÷x/(x-1)
=[(x-1+1)/(x-1)]÷x/(x-1)
=x/(x-1)÷x/(x-1)
=1
(x²-4x+4)/(x²-4)÷(x²-2x)/(x+2)
=(x-2)²/[(x+2)(x-2)]×(x+2)/x(x-2)
=1/x
[a/(ab-b²)-b/(a²-ab)]÷[1+(a²+b²)/2ab]
=[a²/ab(a-b)-b²/ab(a-b)]÷[2ab/2ab+(a²+b²)/2ab]
=[(a+b)(a-b)]/[ab(a-b)]÷(a+b)²/2ab
=(a+b)/ab×2ab/(a+b)²
=2/(a+b)
[(2a-b)/(a+b)-b/(a-b)]÷(a-2b)/(a-b)
={[(2a-b)(a-b)]/[(a+b)(a-b)]-b(a+b)/[(a+b)(a-b)]}×(a-b)/(a-2b)
=(2a²-3ab+b²-ab-b²)/[(a+b)(a-b)]×(a-b)/(a-2b)
=2a(a-2b)/[(a+b)(a-b)]×(a-b)/(a-2b)
=2a/(a+b)
(a²-b²)/(a²-ab)÷[a+(2ab+b²)/a]
=[(a+b)(a-b)]/a(a-b)÷(a²+2ab+b²)/a
=(a+b)/a×a/(a+b)²
=1/(a+b)
计算:(1)(1+1/x-1)÷(x/x-1)(2)x^2-4x+4/x^2-4÷ x^2-2x/x+2(3)(a/ab-b^2-b/a^2-ab)÷(1+a^2+b^2/2ab)
计算:(1)(1+1/x-1)÷(x/x-1)(2)x^2-4x+4/x^2-4÷ x^2-2x/x+2(3)(a/ab-b^2-b/a^2-ab)÷(1+a^2+b^2/2ab)
(4)(2a-b/a+b - b/a-b)÷a-2b/a-b)(5)a^2-b^2/a^2-ab÷(a+2ab+b^2/a)各位大神帮帮忙呢~
(4)(2a-b/a+b - b/a-b)÷a-2b/a-b)(5)a^2-b^2/a^2-ab÷(a+2ab+b^2/a)各位大神帮帮忙呢~
数学人气:633 ℃时间:2020-10-01 20:21:02
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