Question

急急急，怎么做啊？求高手啊！！

Answer 1

1.
原式=5x/(x+3)(x-2)+(2x-5)/(x-4)(x+3)-(7x-10)/(x-2)(x-4)
分母两个两个之间都有共同的项，所以合并分子分母
=[5x(x-4)+(2x-5)(x-2)-(7x-10)(x+3)]/(x+3)(x-2)(x-4)
=(5x^2-20x+2x^2-5x-4x+10-7x^2+10x-21x+30)/(x+3)(x-2)(x-4)
=(-40x+40)/(x+3)(x-2)(x-4)
=40(1-x)/(x+3)(x-2)(x-4)
2.
1/(x+1)(x+2)+1/(x+2)(x+3)=(x+3+x+1)/(x+1)(x+2)(x+3)=2(x+2)/(x+1)(x+2)(x+3)=2/(x+1)(x+3)
2/(x+1)(x+3)+1/(x+3)(x+4)=3(x+3)/(x+1)(x+3)(x+4)=3/(x+1)(x+4)
由此可看出规律
1/(x+1)(x+2)+1/(x+2)(x+3)+...1/(x+n)(x+n+1)=n/(x+1)(x+n+1)
则该题的结果为
100/(x+1)(x+101)

Answer 2

什么怎么做

追问

？，图片上不是在的吗

追答

1.
原式=5x/(x+3)(x-2)+(2x-5)/(x-4)(x+3)-(7x-10)/(x-2)(x-4)
分母两个两个之间都有共同的项，所以合并分子分母
=[5x(x-4)+(2x-5)(x-2)-(7x-10)(x+3)]/(x+3)(x-2)(x-4)
=(5x^2-20x+2x^2-5x-4x+10-7x^2+10x-21x+30)/(x+3)(x-2)(x-4)
=(-40x+40)/(x+3)(x-2)(x-4)
=40(1-x)/(x+3)(x-2)(x-4)
2.
1/(x+1)(x+2)+1/(x+2)(x+3)=(x+3+x+1)/(x+1)(x+2)(x+3)=2(x+2)/(x+1)(x+2)(x+3)=2/(x+1)(x+3)
2/(x+1)(x+3)+1/(x+3)(x+4)=3(x+3)/(x+1)(x+3)(x+4)=3/(x+1)(x+4)
由此可看出规律
1/(x+1)(x+2)+1/(x+2)(x+3)+...1/(x+n)(x+n+1)=n/(x+1)(x+n+1)
则该题的结果为
100/(x+1)(x+101)

追问

谢谢啦，不过迟啦一步，我已经选别人的为满意答案啦..

追答

额

追问

没办法啦