cho : `x;y≥0` `x+y=1` tìm `maxA=(x)/(y+1)+(y)/(x+1)` 06/07/2021 Bởi Caroline cho : `x;y≥0` `x+y=1` tìm `maxA=(x)/(y+1)+(y)/(x+1)`
`A=(x)/(y+1)+y/(x+1)=(y(y+1)+(x+1)x)/((y+1)(x+1))=(y^2+x^2+x+y)/(xy+1+x+y)=((x+y)^2-2xy+1)/(xy+2)=(-2xy+2)/(xy+2)=-2+(6)/(xy+2)≤-2+(6)/(2)≤-2+6/2≤1` `”=”`xẩy ra khi :`x=0;y=1` hoặc `x=1;y=0` Bình luận
`A=(x)/(y+1)+y/(x+1)=(y(y+1)+(x+1)x)/((y+1)(x+1))=(y^2+x^2+x+y)/(xy+1+x+y)=((x+y)^2-2xy+1)/(xy+2)=(-2xy+2)/(xy+2)=-2+(6)/(xy+2)≤-2+(6)/(2)≤-2+6/2≤1`
`”=”`xẩy ra khi :
`x=0;y=1` hoặc
`x=1;y=0`