Question

方程 5/(x-2)+3/(2-x)=2 计算 (3x-2)/(x^2-x-2)+[1-1/(x+1)]/[1+1/(x-1)]

Answer 1

方程 5/(x-2)+3/(2-x)=2
两边乘以x-2得
5-3=2(x-2)
∴x=1
经检验：x=1是方程的解
∴方程的解是x=1

计算 (3x-2)/(x^2-x-2)+[1-1/(x+1)]/[1+1/(x-1)]
=(3x-2)/[(x-2)(x+1)]+[x/(x+1)]/[x/(x-1)]
=(3x-2(/[(x-2)(x+1)]+(x-1)/(x+1)
=[(3x-2)+(x-2)(x-1)]/[(x-2)(x+1)]
=x²/(x²-x-2)

Answer 2

方程 5/(x-2)+3/(2-x)=2
两边乘以x-2得
5-3=2(x-2)
2=2x-4
2x=6
x=3
(3x-2)/(x^2-x-2)+[1-1/(x+1)]/[1+1/(x-1)]

= (3x-2)/[(x+1)(x-2)]+(x-1)/(x+1)
=(3x-2)/[(x+1)(x-2)]+(x-1)(x-2)/(x+1)(x-2)
=x²/(x²-x-2)
将x=3代入，得
3²/(3²-3-2)
=9/4

Answer 3

5/(x-2)+3/(2-x)=2
2/(x-2)=2
x=3带入(3x-2)/(x^2-x-2)+[1-1/(x+1)]/[1+1/(x-1)]