求y=x+根号下x(2-x) 的值域

y=x+√[x(2-x)]
定义域为x(2-x)≤0 ===> x(x-2)≤0 ===> 0≤x≤2
原式===> y-x=√[x(2-x)]
===> (y-x)^2=x(2-x)=2x-x^2
===> y^2-2yx+x^2-2x+x^2=0
===> 2x^2-2(y+1)x+y^2=0
由△=4(y+1)^2-8y^2≥0 ===> (y+1)^2-2y^2≥0
===> y^2+2y+1-2y^2≥0
===> y^2-2y-1≤0
===> (2-2√2)/2≤y≤(2+2√2)/2
===> 1-√2≤y≤1+√2
而0≤x≤2时,x≥0,√[x(2-x)]≥0
所以,y≥0
综上：y∈[0,1+√2]

1年前