求函数y=(2x^2-8x+16)/(x^2-2x+4) (x>2)的值域.用均值不等式解

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求函数y=(2x^2-8x+16)/(x^2-2x+4) (x>2)的值域.用均值不等式解

y=(2x^2-8x+16)/(x^2-2x+4)
=[2(x^2-2x+4)-4x+8]/(x^2-2x+4)
=2+[(8-4x)/(x^2-2x+4)]
余下省略.

y=(2x^2-8x+16)/(x^2-2x+4)
=2+(8-4x)/(x^2-2x+4)
=2+4/[(x^2-2x+4)/(2-x)]
=2-4/{[x(x-2)+4]/(x-2)}
=2-4/[x+4/(x-2)]
=2-4/[x-2+2+4/(x-2)]
=2-4/[x-2+4/(x-2)+2]
>=2-4/{[(2√(x-2)*4/(x-2)]+2}
=4/3
故值域为【2/3,+∞】