求值域
求u=(cost-2sint+1)/(2cost+sint-1)的值域.
解: u=(cost-2sint+1)/(2cost+sint-1) --->(u+2)sint+(2u-1)cost=u+1 故由Cauchy不等式,得 [(u+2)^2+(2u-1)^2][(sint)^2+(cost)^2]>=[(u+2)sint+(2u-1)cost]^2 --->(u+2)^2+(2u-1)^2>=(u+1)^2 --->2(u-1/4)^2+15/8>=0 u为任何实数时,上式均成立. 故u值域为R.
答:求函数y=√(x^2+x+1)-√(x^2-x+1)的值域 y=√(x^2+x+1)-√(x^2-x+1) =√[(x+1/2)^2+(√3/2)^2]-√[(...详情>>
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