Question

y=2x的平方-x+2\x的平方+x+1求值域

Answer 1

y=(2x^2-x+2)/(x^2+x+1)
y=(2x^2+2x+2-3x)/(x^2+x+1)
y=2-3x/(x^2+x+1)
令3x/(x^2+x+1)=t
1/t=(x^2+x+1)/(3x)
1/t=(1/3)(x+1+1/x)=(1/3)(x+1/x)+1/3
利用均值不等式x+1/x>=2
所以1/t>=1
0<=t<=1
1<=y<=2

Answer 2

1.y=(2x^2-x+2)/(x^2+x+1)
(x²+x+1)y=2x²-x+2
(y-2)x²+(y+1)x+y-2=0
△=(y+1)²-4(y-2)²≥0
=>(y-5)(3y-3)≤0
=>1≤y≤5