Question

初中两道分式计算题~求高人解：）

1. 已知2x-3/(x-1)(x+2)=A/x-1+B/x+2，则A= ，B= 。
2.计算：2/(x+1)(x+3)+2/(x+3)(x+5)…+2/(x+1997)(x+1999)

一小题我得后一步是A(x+2)+B(x-1)=2x-3，但是解不出来呢？
然后第二题怎么做？做第二题这种类型的题应该怎么着手呢？

Answer 1

2x-3/(x-1)(x+2)=A/x-1+B/x+2，则A= ，B= 。
2x-3/(x-1)(x+2) =A (x+2)/(x-1) (x+2) +B(x-1) /(x-1)(x+2)
2x-3/(x-1)(x+2) =A (x+2) +B(x-1) /(x-1)(x+2)
方程两边同乘(x-1)(x+2)
2x-3=A (x+2) +B(x-1)
2x-3=Ax+2A+BX-B
2x-3=X(A+B)+2A-B
所以A+B=2； 2A-B=-3
解二元一次方程组，得A=-1/3 B=7/3

2/(x+1)(x+3)+2/(x+3)(x+5)…+2/(x+1997)(x+1999)
=1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+......+1(x+1997)-1/(x+1999)
=1/(x+1)-1/(x+1999)
=1998/(x+1)(x+1999)

Answer 2

1、A/x-1+B/x+2
=【A(x+2)+B(x-1)】/(x-1)(x+2)=2x-3)/(x-1)(x-2)
A(x+2)+B(x-1)=2x-3
Ax+2A+Bx-B=2x-3
x(A+B)+(2A-B)=2x-3
A+B=2
2A-B=-3
A=-1/3
B=7/3
2、2/(x+1)(x+3)+2/(x+3)(x+5)…+2/(x+1997)(x+1999)
=1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+......+1(x+1997)-1/(x+1999)
=1/(x+1)-1/(x+1999)
=1998/(x+1)(x+1999)

Answer 3

A=5/3,B=1/3 第一题：先把等号右面的通分，变为A（x+2)+B(x-1)/(x-1)(x-2) 然后开始整理，把有x的变为一项：x(A+B)+2A-B/(x-1)(x+2) 这时再看下等号左边的，列一个二元一次方程，就能解出来

Answer 4

2x-3/(x-1)(x+2)=A/x-1+B/x+2，则A= ，B= 。
2x-3/(x-1)(x+2) =A (x+2)/(x-1) (x+2) +B(x-1) /(x-1)(x+2)
2x-3/(x-1)(x+2) =A (x+2) +B(x-1) /(x-1)(x+2)
方程两边同乘(x-1)(x+2)
2x-3=A (x+2) +B(x-1)
2x-3=Ax+2A+BX-B
2x-3=X(A+B)+2A-B
所以A+B=2； 2A-B=-3解二元一次方程组，得A=-1/3 B=7/3
2/(x+1)(x+3)+......2/(x+1997)(x+1999)
=1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+......+1(x+1997)-1/(x+1999)
=1/(x+1)-1/(x+1999)
=1998/(x+1)(x+1999)

Answer 5

(1)A(x+2)+B(x-1)=2x-3
Ax+2A+Bx-B=2x-3
(A+B)x+(2A-B)=2x-3
就有A+B=2 ，2A-B=-3
解此方程组得A=-1/3，B=7/3
（2）有公式
例：1/1*2+1/2*3+.......+1/(n-1)n+1/n(n+1)
=1/1[1/1-1/2+1/2-1/3+......+1/(n-1)-1/n+1/n-1/(n+1)]
=1-1/(n+1)
1/1*3+1/3*5+.......+1/(n-2)n+1/n(n+2)
=1/2[1/1-1/3+1/3-1/5+.........+1/(n-2)-1/n+1/n-1/(n+2)]
=1/2[1-1/(n+2)]