（1）求{x|x=cos(kπ/2),k∈Z}∩{x|x=sin(2kπ±π/2),k∈Z}
2个回答

(1)

{x|x=cos(kπ/2),k∈Z} = {1,0,-1}

{x|x=sin(2kπ±π/2),k∈Z} = {1,-1}

{x|x=cos(kπ/2),k∈Z}∩{x|x=sin(2kπ±π/2),k∈Z} = {1,-1}

(2)即 x^2 + mx + 4 = 0无实数解,即 m^2 - 16 < 0,即m = 0,1,2,3

{(x,y)|(x-y)根号x=0} = {直线x=y和直线x = 0}

{(x,y)||y|=1}={直线y=1和直线y=-1}

{(x,y)|(x-y)根号x=0}∩{(x,y)||y|=1}={(-1,-1),(0,-1),(0,1),(1,1)}

另：楼主高中文科么?这个应该都不难的

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