Question

设a={x/x^2+2(a+1)x+a^2-1=0},b={x/(x+4)x(x- 1/2） =0 ,x属于z},若a交

Answer 1

B = {x|(x+4)x(x-1/2)=0,x∈Z} = {0,-4}

A∩B=A--->A = "空集"或{0}或{-4}或B

（1）A = "空集"

--->二次方程x²+2(a+1)x+a²-1=0无解

--->Δ = 4(a+1)²-4(a²-1) = 4(2a+2)＜0--->a＜-1

（2）A = {0}或{-4}

--->二次方程x²+2(a+1)x+a²-1=0有重根0或-4

--->Δ=0--->a=-1--->A={0}

（3）A = B = {0,-4}

--->二次方程x²+2(a+1)x+a²-1=0有两根0、-4

--->0-4=-2(a+1)且0•(-4)=a²-1--->a=1

综上：a∈(-∞,-1]∪{1}