Question

解分式方程度：[(2x-5)/(x^2+3x+2)]+[4/(x^2-4)]=[2/(x-2)]

要过程！！！一定要清楚！！！

Answer 1

[(2x-5)/(x^2+3x+2)]+[4/(x^2-4)]=[2/(x-2)]
[(2x-5)/(x+1)(x+2)]+[4/(x+2)(x-2)]=[2/(x-2)] (分母因式分解)
(2x-5)(x-2)+4(x+1)=2(x+1)(x+2) (两边乘以(x+1)(x+2)(x-2))
2x^2-9x+10+4x+4=2x^2+6x+4 (乘开)
11x=10 (整理)
x=10/11

经检验，X=10/11是原方程的解

Answer 2

(2X-5)/[(x+1)(X+2)]+4/[(x-2)(x+2)]=2/(X-2)
两边乘以(x-2)(x+2)(x+1)
(2X-5)(X-2)+4(X+1)=2(X+1)(X+2)
2X^2-9X+10+4X+4=2X^2+6X+4
2X^2-5X+14=2X^2+6X+4
11X=10
X=10/11
经检验，X=10/11是原方程的解

Answer 3

[(2x-5)/(x^2+3x+2)]+[4/(x^2-4)]=[2/(x-2)]
(2x-5)/[(x+2)(x+1)]+4/(x+2)(x-2)=2/(x-2)(两边都乘以(x-2)(x+2)(x+1))
(2x-5)(x-2)+4(x+1)=2(x+2)(x+1)
(2x^2-7x+10)+(4x+4)=2(x^2+3x+2)
2x^2-3x+14=2x^2+6x+4
-9x+10=0
x=10/9

Answer 4

(2x-5)/[(x+1)(x+2)] + 4/[(x+2)(x-2)]
=2(x+2)/[(x+2)(x-2)]

(2x-5)/[(x+1)(x+2)] + 4/[(x+2)(x-2)]
=(2x+4)/[(x+2)(x-2)]

(2x-5)/[(x+1)(x+2)]=2x/[(x+2)(x-2)]

(2x-5)/(x+1)=2x/(x-2)

(2x-5)(x-2)=2x(x+1)

2x^2-9x+10=2x^2+2x

11x=10

x=10/11

又因为10/11!=-1 10/11!=2 且10/11!=-2
所以x=10/11是原方程的解

Answer 5

[(2X-5)/(X+2)(X+1)]+[4/(X+2)(X-2)]=[2/[X-2]]

两边都c乘以(x+1)(x+2)(x-2)后得出：

(2x-5)(x-2) + 4(x+1) = 2(x+1)(x+2)

2X^2-9X+10+4X+4=2X^2+6X+4

11X=10

X=10/11

Answer 6

两边同乘以(x+1)(x+2)(x-2)得
(2x-5)(x-2) + 4(x+1) = 2(x+1)(x+2)
==>x = 10/11;