1.D 2.C 3.C 4.C 5.-3
6.t₁=0,t₂=-3/2
7.解:(1)由2x²=32,
∴x²=16,
∴x₁=4,x₂=-4
(2)t₁=2,t₂=3
(3)(y-1)(3y-2)=0 ∴y₁=2/3,y₂=1
(4)x₁=9/2,x₂=-1/2
(5)(3x+2)[(3x+2)-(x-1)]=0 (3x+2)(5x+5)=0,∴x₁=-2/3,x₂=-1
(6)方法一:直接开平方,得3x-1=±(3-2x)
即3x-1=3-2x或3x-1=-3+2x,
∴x₁=4/5,x₂=-2.
8.解:小亮的解法正确而且简便
小明的解法违反了等式的基本性质,应分2x+3=0和2x+3≠0两种情况讨论.
9.解:(1)B (2)换元法 (3)完整
(4)①t₁=2,t₂=-2,t₃=3,t₄=-3;②x₁=0,x₂=1 (5)A