Question

1/(X+2)(X+3)+1/(X+3)(X+4)=??怎么算，详细，谢谢

Answer 1

解：1/（x+1）+1/（x+5）=1/（x+2）+1/（x+4）
移项：1/（x+1）-1/（x+4）=1/（x+2)-1/（x+5）
通分：3/（x+1）（x+4）=3/（x+2）（x+5）
所以（x+1）（x+4）=（x+2）（x+5）
即：x²+5x+4=x²+7x+10
所以2x=-6
x=-3
遇到题目的时候先观察，不要盲目计算哦！
谢谢采纳！

Answer 2

1/(X+2)(X+3)+1/(X+3)(X+4)
=1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)
=1/(x+2)-1/(x+4)
=(x+4-x-2)/(x+2)(x+4)
=2/(x+2)(x+4);

如果本题有什么不明白可以追问，如果满意记得采纳
如果有其他问题请采纳本题后另发点击向我求助，答题不易，请谅解，谢谢。
祝学习进步

追问

1/（x+2008)(x+2009)=??????

追答

1/（x+2008)(x+2009)
=1/(x+2008)-1/(x+2009);

Answer 3

如有帮到你还望采纳，谢谢

Answer 4

通分
1/(X+2)(X+3)+1/(X+3)(X+4)
=1*（X+4)/(X+2)(X+3)(X+4)+1*(X+2)/(X+3)(X+4)(X+2)
=[1*（X+4)+1*(X+2)]/(X+3)(X+4)(X+2)
=(2X+6)/(X+3)(X+4)(X+2)
=2(X+3)/(X+3)(X+4)(X+2)
=2/(X+4)(X+2)

Answer 5

1/(X+2)(X+3)+1/(X+3)(X+4)
=[1/(X+2)][1/(X+3)]+[1/(X+3)][1/(X+4)]
=[1/(X+3)][1/(X+2)+1/(X+4)]
=[1/(X+3)][(X+2)+(X+4)]/(X+2)(X+4)]
=[1/(X+3)][(2X+6)/(X+2)(X+4)]
=[1/(X+3)][2(X+3)/(X+2)(X+4)]
=2/(X+2)(X+4)
=[(X+4)-(X+2)]/(X+2)(X+4)
=[(X+4)/(X+2)(X+4)]-[(X+2)/(X+2)(X+4)]
=[1/(X+2)]-[1/(X+4)]