(1) 2lg(x-1)≥lg(7-x)
lg[(x-1)^2]≥lg(7-x)
(x-1)^2≥(7-x)
x^2-x-6≥0
x≤-2 or x≥3
另由于lg函数定义域的限制,有
x-1≥0 and 7-x≥0
故 3≤x≤7
(2)先求a,b内积,并记a,b夹角为α
因(3a-b)⊥(2a+3b) 有
=0
6-2+9+3=0
6|a|^2+7-3|b|^2=0
=6/7
从而 cos α = /(|a||b|)=3/7
α=arccos 3/7
(3)根号3记作sqr(3),则由两直线平行知
sqr(3)/(-3)=sinα/cosβ
α = arcsin [-cosβ*sqr(3)/3]
然后根据0≤α≤π 作调整
p.s. 在下怀疑本题是否有问题,若直线二为xsinα+ycosα-1=0 则可解出唯一α = 5π/6
(4)A={x/-2≤x≤2,x∈z}= {-2,-1,0,1,2}
B={y/y=x*x,x∈A}= {0,1,4}
A∪B = {-2,-1,0,1,2,4}