已知m={m|方程mx²-x-1=0有实数根},N={n|方程x²-x+n=0有实数根},求M∪N M∩N

已知m={m|方程mx²-x-1=0有实数根},N={n|方程x²-x+n=0有实数根},求M∪N M∩N
已知m={m|方程mx²-x-1=0有实数根},N={n|方程x²-x+n=0有实数根},求M∪N M∩N

已知m={m|方程mx²-x-1=0有实数根},N={n|方程x²-x+n=0有实数根},求M∪N M∩N
M={|mx²-x-1=0有实数根},则方程mx²-x-1=0的△≥0
即△=(-1)²-4m(-1)=1+4m≥0
所以m≥-1/4
CuM={m|m<-1/4}
N={n|x²-x+n=0有实数根},则方程x²-x+n=0的△≥0
即△=(-1)²-4n=1-4n≥0
n≤1/4
N={n|n≤1/4}
(CuM)∩N={m|m<-1/4}∩{n|n≤1/4}=(-∞,-1/4)
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