解不等式组-1≤x-1≤1,-1≤x^2-1≤1,x-1≤x^2-1,x的取值范围是?

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解不等式组-1≤x-1≤1,-1≤x^2-1≤1,x-1≤x^2-1,x的取值范围是?
解不等式组-1≤x-1≤1,-1≤x^2-1≤1,x-1≤x^2-1,x的取值范围是?

解不等式组-1≤x-1≤1,-1≤x^2-1≤1,x-1≤x^2-1,x的取值范围是?
-1≤x-1≤1的解为0≤x≤2
-1≤x^2-1≤1两边同时加一,得出:-1≤x≤1
x-1≤x^2-1,移项:
x^2-x≥0,解得:x≥1或x≤0
综上所述
x=1或0

由-1≤x-1≤1得0≤x≤2;
由-1≤x^2-1≤1得-根号2≤x≤根号2;
由x-1≤x^2-1得x≤0或x≥1
故不等式组的解是{x|1≤x≤根号2或x=0}