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
1个回答

/(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)

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