Question

超级简单的分式加减运算！高手进来看看！要详细解答过程！

1、a-1分之2+2a+1分之a+a²+a-2分之a²-4

2、（1+x）（x²+x分之x-1 - x³-x分之1-x）

3、（a-1分之a²-1 + 1）/（a+3）*a²+2a分之a²-1

Answer 1

1、
原式=2/(a-1)+a/(2a+1)+(a+2)(a-2)/(a+2)(a-1)
=2/(a-1)+a/(2a+1)+(a-2)/(a-1)
=(2+a-2)/(a-1)+a/(2a+1)
=[a(2a+1)+a(a-1)]/(a-1)(2a+1)
=(2a^2+a+a^2-a)/(a-1)(2a+1)
=3a^2/(a-1)(2a+1)

2、
原式=(1+x)[(x-1)/x(1+x)-(1-x)/x(x+1)(x-1)]
=(1+x)*(x-1)/x(1+x)-(1+x)*(1-x)/x(x+1)(x-1)
=(1+x)*(x-1)/x(1+x)+(1+x)*(x-1)/x(x+1)(x-1)
=(x-1)/x+1/x
=1-1/x+1/x
=1

3、
原式=[(a+1)(a-1)/(a-1)+1]/(a+3)*[(a+1)(a-1)/a(a+2)]
=(a+2)(a+1)(a-1)/[(a+3)a(a+2)]
=(a+1)(a-1)/[a(a+3)]
=(a^2-1)/(a^2+3a)