解由x2 +2y2=6
得x2 /6+y2/3=1
故设x=√6cosa,y=√3sina
则x+y=√6cosa+√3sina
=3(√6/3cosa+√3/3sina)
=3sin(a+θ)
由-3≤3sin(a+θ)≤3
即-3≤x+y≤3
故x+y的范围是[-3,3].
设实数xy满足x2 +2y2=6,则x+y的取值范围
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