作业帮计算 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
来源：学生作业帮 编辑：拍题作业网作业帮 分类：数学作业 时间：2024/04/26 23:47:52
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
/(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)
分式习题(1)式子A/BC + B/CA + C/AB的值能否为0?为什么?(2)式子A-B/(B-C)(C-A) +
（1）式子a/bc+b/ca+c/ab的值能否为0?为什么?（2）式子a-b|（b-c)(c-a)+b-c|（a-b)(
ab+a+b=bc+b+c=ca+c+a=3,求(a+1)(b+1)(c+1)得多少,如何计算的,
计算行列式D={1 a bc}{1 b ca}{1 c ab}的值 D=(b-a)(b-c)(a-c)怎么算
已知a×a+b×b=1,b×b+c×c=2,c×c+a×a=2,求ab+bc+ca的最小值是多少?
初二分式的加减法2a-b-c/（a-b)(a-c)+ 2b-c-a/(b-c)(b-a)+ 2c-b-a/(c-b)(c
已知a>b>c,求证a²b+b²c+c²a>ab²+bc²+ca
计算： ab/(b-c)(c-a)+bc/(a-b)(c-a)+ac/(a-b)(b-c)
已知a+b+c=1求证ab+bc+ca
因式分解abc+ab+bc+ca+a+b+c+1=
(ab+bc+ca+1+a+b+c+abc)(ab+bc+ca+1-a-b-c-abc)怎么化简?
用分式加减法计算：b/a-b + a/b-a