设x=sina(a∈[-π/2,π/2]),方程变为cosa=sina+m,
所以m=cosa-sina=-√2sin(a-π/4)
∵a∈[-π/2,π/2],∴a-π/4∈[-3π/4,π/4],
∴sin(a-π/4)∈[-1,√2/2],∴y∈[-√2,1]
因为方程无解,所以m1
若方程根号下1-x的平方=x+m无实数解,则实数m 的取值范围是 +解析
若方程根号下1-x的平方=x+m无实数解,则实数m 的取值范围是 +解析
数学人气:478 ℃时间:2019-11-16 11:43:08
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