已知x、y为正数,且x2+y2/2=1,则x√(1+y2)的最大值为?,x=?已知x,y∈R,且(x^2+y^2)/2=1,则 x√(1+y^2)的最大值∵(x^2+y^2)/2=1,∴x^2+y^2=2x√(1+y^2)= √[x^2(1+y^2)≤(1/2)[x^2+(1+y^2)]=(1/2)(2+1)=3/2∴最大值为3/2我是这样做
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![已知x、y为正数,且x2+y2/2=1,则x√(1+y2)的最大值为?,x=?已知x,y∈R,且(x^2+y^2)/2=1,则 x√(1+y^2)的最大值∵(x^2+y^2)/2=1,∴x^2+y^2=2x√(1+y^2)= √[x^2(1+y^2)≤(1/2)[x^2+(1+y^2)]=(1/2)(2+1)=3/2∴最大值为3/2我是这样做](/uploads/image/z/2624413-13-3.jpg?t=%E5%B7%B2%E7%9F%A5x%E3%80%81y%E4%B8%BA%E6%AD%A3%E6%95%B0%2C%E4%B8%94x2%2By2%2F2%3D1%2C%E5%88%99x%E2%88%9A%281%2By2%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA%3F%2Cx%3D%3F%E5%B7%B2%E7%9F%A5x%2Cy%E2%88%88R%2C%E4%B8%94%28x%5E2%2By%5E2%29%2F2%3D1%2C%E5%88%99+x%E2%88%9A%281%2By%5E2%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E2%88%B5%28x%5E2%2By%5E2%29%2F2%3D1%2C%E2%88%B4x%5E2%2By%5E2%3D2x%E2%88%9A%281%2By%5E2%29%3D+%E2%88%9A%5Bx%5E2%281%2By%5E2%29%E2%89%A4%281%2F2%29%5Bx%5E2%2B%281%2By%5E2%29%5D%3D%281%2F2%29%282%2B1%29%3D3%2F2%E2%88%B4%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA3%2F2%E6%88%91%E6%98%AF%E8%BF%99%E6%A0%B7%E5%81%9A)
已知x、y为正数,且x2+y2/2=1,则x√(1+y2)的最大值为?,x=?已知x,y∈R,且(x^2+y^2)/2=1,则 x√(1+y^2)的最大值∵(x^2+y^2)/2=1,∴x^2+y^2=2x√(1+y^2)= √[x^2(1+y^2)≤(1/2)[x^2+(1+y^2)]=(1/2)(2+1)=3/2∴最大值为3/2我是这样做
已知x、y为正数,且x2+y2/2=1,则x√(1+y2)的最大值为?,x=?
已知x,y∈R,且(x^2+y^2)/2=1,则 x√(1+y^2)的最大值
∵(x^2+y^2)/2=1,∴x^2+y^2=2
x√(1+y^2)= √[x^2(1+y^2)
≤(1/2)[x^2+(1+y^2)]=(1/2)(2+1)=3/2
∴最大值为3/2
我是这样做的但是错了,答案是3根号2/4,后来老师说要乘系数,变成x√(1+y^2)= √[1/2*(2x^2)(1+y^2)才能做对.求解为什么要乘系数?其他题我那样做能做对为什么这道题不行
打错了
∵x^2+y^2/2=1,∴2x^2+y^2=2
x√(1+y^2)= √[x^2(1+y^2)
≤(1/2)[x^2+(1+y^2)]=(1/2)(2+1)=3/2
∴最大值为3/2
已知x、y为正数,且x2+y2/2=1,则x√(1+y2)的最大值为?,x=?已知x,y∈R,且(x^2+y^2)/2=1,则 x√(1+y^2)的最大值∵(x^2+y^2)/2=1,∴x^2+y^2=2x√(1+y^2)= √[x^2(1+y^2)≤(1/2)[x^2+(1+y^2)]=(1/2)(2+1)=3/2∴最大值为3/2我是这样做
第一种方法:
令x=cost,y=根号2sint t∈[0,π/2]
则x√(1+y2)=cost根号(1+2sin²t)=根号(cos²t+2sin²tcos²t)
=根号(cos²t-1/2+1/2sin²2t+1/2)=根号(1/2cos2t+1/2-1/2cos²2t+1/2)
=根号(-1/2cos²2t+1/2cos2t+1)=根号[-1/2(cos2t-1/2)²+9/8]
∵t∈[0,π/2]∴cos2t∈[-1,1]
所以当cos2t=1/2时最大,此时x=cost=cosπ/6=根号3/2,得x√(1+y2)最大值=根号9/8=3根号2/4
第二种方法1+y²=3-2x²
则x√(1+y2)=x根号(3-2x²)=根号(3x²-2x四次方)
令t=x²,因为x、y为正数,且x2+y2/2=1所以x∈[0,1]
∴t∈[0,1]
根号(3x²-2x四次方)=根号(3t-2t²)=根号[-2(t-3/4)²+9/8]
当t=3/4时最大,此时x=根号t=根号3/2,则x√(1+y2)最大值=根号9/8=3根号2/4
第三种方法1+y²=3-2x²
则x√(1+y2)=x根号(3-2x²)=根号2[x(3/2-x)]利用均值不等式最大值可求