x属于(0,+无穷),x²+2y²=1,求M=x根号(1+y²)的最大值
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x属于(0,+无穷),x²+2y²=1,求M=x根号(1+y²)的最大值
x属于(0,+无穷),x²+2y²=1,求M=x根号(1+y²)的最大值
x属于(0,+无穷),x²+2y²=1,求M=x根号(1+y²)的最大值
答:
x>0,x²+2y²=1
M=x√(1+y²)
=x√[1+(1-x²)/2]
=x√(3/2-x²/2)
=(√2/2) x√(3-x²)
>0
所以:0