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

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

人气：158 ℃　时间：2020-03-26 10:15:51

解答

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...