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计算分式的加减法.b/(a+b)-a/a-b(a-b/ab)+(b-c/bc)+(c-a/ca)(1/x-1)-(1/x+1)-(2/x^2+1)(4/a+2)+a-2
题目详情
计算 分式的加减法.
b/(a+b)-a/a-b
(a-b/ab)+(b-c/bc)+(c-a/ca)
(1/x-1)-(1/x+1)-(2/x^2+1)
(4/a+2)+a-2
b/(a+b)-a/a-b
(a-b/ab)+(b-c/bc)+(c-a/ca)
(1/x-1)-(1/x+1)-(2/x^2+1)
(4/a+2)+a-2
▼优质解答
答案和解析
b/(a+b)-a/(a-b)
=[b(a-b)-a(a+b)]/(a²-b²)
=(-b²-a²)/(a²-b²)
=(b²+a²)/(b²-a²)
(a-b/ab)+(b-c/bc)+(c-a/ca)
=(1/b)-(1/a)+(1/c)-(1/b)+(1/a)-(1/c)
=0
(1/x-1)-(1/x+1)-(2/x²+1)
=[(x+1)-(x-1)]/(x²-1)-2/(x²+1)
=2/(x²-1)-2/(x²+1)
=2[(x²+1)-(x²-1)]/(x^4-1)
=4/(x^4-1)
(4/a+2)+a-2
=4/(a+2)+(a²-4)/(a+2)
=[4+(a²-4)]/(a+2)
=a²/(a+2)
=[b(a-b)-a(a+b)]/(a²-b²)
=(-b²-a²)/(a²-b²)
=(b²+a²)/(b²-a²)
(a-b/ab)+(b-c/bc)+(c-a/ca)
=(1/b)-(1/a)+(1/c)-(1/b)+(1/a)-(1/c)
=0
(1/x-1)-(1/x+1)-(2/x²+1)
=[(x+1)-(x-1)]/(x²-1)-2/(x²+1)
=2/(x²-1)-2/(x²+1)
=2[(x²+1)-(x²-1)]/(x^4-1)
=4/(x^4-1)
(4/a+2)+a-2
=4/(a+2)+(a²-4)/(a+2)
=[4+(a²-4)]/(a+2)
=a²/(a+2)
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