(1)解不等式2lg(x-1)≥lg(7-x)

1个回答

  • (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}