38.3.已知集合M={y | y = x2},N={y | x2+y2=2},则M∩N=( D )A.{(1,1),(-1,1)} B.{1} C.[0,1] D.[0,√2]

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38.3.已知集合M={y | y = x2},N={y | x2+y2=2},则M∩N=( D )A.{(1,1),(-1,1)} B.{1} C.[0,1] D.[0,√2]
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38.3.已知集合M={y | y = x2},N={y | x2+y2=2},则M∩N=( D )A.{(1,1),(-1,1)} B.{1} C.[0,1] D.[0,√2]
38.3.已知集合M={y | y = x2},N={y | x2+y2=2},则M∩N=( D )
A.{(1,1),(-1,1)} B.{1} C.[0,1] D.[0,√2]

38.3.已知集合M={y | y = x2},N={y | x2+y2=2},则M∩N=( D )A.{(1,1),(-1,1)} B.{1} C.[0,1] D.[0,√2]
My=x²≥0
在x²+y²=2中
y=+/-√2-x²
因为x²≥0
所以-x²≤0
所以y∈【-√2,√2】
所以所以M∩N={x|0≤x≤√2 }
(哎~一楼抄答案抄的太快乐 连等号都漏了╮(╯_╰)╭

M={y|y>=0}
N={y|-根2<=y<=根2}
所以M∩N={y|0<=y<=根2 }

M={y|y>=0}
N={y|-根2<=y<=根2}
所以M∩N={x|0