解一道不等式。谢谢

解:由题意,
x^2/(3-2x)≥1
x^2/(3-2x)-1>=0
[x^2-(3-2x)]/(3-2x)>=0
(x+3)(x-1)(3-2x)>=0
(x+3)(x-1)(2x-3)<=0

所以x<=-3或
1 <= x <=3/2.

x^2/（3-2x）-1=0的两根是 -3 和 1

x>=1或者x<=-3

x^2≥3-2x
x^2+2x-3≥0
(x+3)(x-2)≥0
x≥-3,x≥2 2大于取大
答案是X≥2

x^2/(3-2x)≥1
x^2/(3-2x)-1>=0
[x^2-(3-2x)]/(3-2x)>=0
(x+3)(x-1)(3-2x)>=0
(x+3)(x-1)(2x-3)<=0

x<=-3或1<=x<3/2,

由题：x^2≥3-2x
x^2+2x-3≥0
(x+3)(x-1)≥0

所以x≥1,或x<=-3

负无穷大到3-2更好下3并上3-2更好下3到无穷大