Question

先化简，再求值：(x^2-4)/(x^2-4x+4)÷(x+2)/(x+1)-x/(x-2)，其中x=2-根号2

要有过程！

Answer 1

解：原式=（X+2）（X-2）/（X-2）²÷（X+2）/（X+1）-X/（X-2）
=（X+2）/（X-2）×（X+1）/（X+2）-X/（X-2）
=（X+1）/（X-2）-X/（X-2）
=1/（X-2）
代入X=2-√2
1/（-√2）=-√2/2

Answer 2

(x^2-4)/(x^2-4x+4)÷(x+2)/(x+1)-x/(x-2)
=(x+2)(x-2)/(x-2)^2÷(x+2)/(x+1)-x/(x-2)
=[(x+2)/(x-2)][(x+1)/(x+2)]-x/(x-2)
=(x+1)/(x-2)-x/(x-2)
=1/(x-2)
=1/(2-√2-2)
=-1/√2
=-√2/2

Answer 3

原式=(x+2)(x-2)/(x-2)^2×(x+1)/(x+2)-x/(x-2)
=(x+2)/(x-2)×(x+1)/(x+2)-x/(x-2)
=(x+1)/(x-2)-x/(x-2)
=(x+1-x)/(x-2)
=1/(x-2)
=1/(2-√2-2)
=-√2/2

Answer 4

=(x+2)(x-2)/[(x-2)(x-2)]÷(x+2)/(x+1)-x/(x-2)
=[(x+2)/(x-2)]×(x+1)/(x+2)-x/(x-2)
=(x+1)/(x-2)-x/(x-2)
=1/(x-2)
代入x
=1/-根号2
＝－根号2/2

Answer 5

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