已知a与b互为相反数,c与d互为倒数,x=5（a-1)-(3a-2b-2),y=cd^2+c^2-(c/d+d-2)

a=-b cd=1
(1) x=5(a-1)-(5a-2)=-3 y=d+c^2-c^2-d+2=2
(2)原式=3-3(1/4-1/5+1/5-1/6+1/6-1/7+.+1/2009-1/2010)=3-3(1/4-1/2010)=3(3/4+1/2010) =753/635我是这样算的Xx=5a-5-3a+2b+2 =2a+2b-3 =-3我刚刚算错了。。。。哦，y的过程能不能详细一点啊y=cd^2+c^2-(c/d+d-2)=cd*d+c^2-(c*c+d-2）=d+c^2-c^2-d+2=2c=1/d cd=1请问第二题是如何变形的呢？麻烦讲解一下。谢谢因为1/n(n+1)=1/n-1/(n+1) 1/(n+1)(n+2)=1/(n+1)-1/(n+2)1/n(n+1+1/(n+1)(n+2)=1/n-1/(n+2)原式后半部分提取x作为公因式然后剩余部分为 1/4*5+1/5*6+1/6*7+.....+1/2009*2010=1/4-1/5+1/5-1/6+1/6-1/7+......+1/2009-1/2010（2）3/4-3(1/4-1/5+1/5-1/6+1/6-1/7+......+1/2009-1/2010)=3/4-3(1/4-1/2010)=3/2010=1/670 刚刚把前面那个看做是（3/2)y了