tìm giá trị nhỏ nhất A=(x-2019)^2+(x+2020)^2 B=x^2 / (y-1) + y^2/(x-1) và (x,y>1) 17/10/2021 Bởi Kylie tìm giá trị nhỏ nhất A=(x-2019)^2+(x+2020)^2 B=x^2 / (y-1) + y^2/(x-1) và (x,y>1)
`A=(x-2019)^2+(x+2020)^2` `⇔A=x^2-4038x+2019^2+x^2+4040x+2020^2` `⇔A=2x^2+2019^2+2x+2020^2` `⇔A=2x^2+2019^2+2x+2020^2` `⇔A=2(x^2+x+1/4)+2019^2-1/2+2020^2` `⇔A=2(x+1/2)^2+2019^2-1/2+2020^2≥2019^2-1/2+2020^2` `”= xẩy ra khi”` `x+1/2=0` `⇒x=-1/2` `⇒minA=2019^2-1/2+2020^2 ` khi` x=-1/2` `B=x^2 / (y-1) + y^2/(x-1)` theo `co-si` `x^2 / (y-1)+4(y-1)≥2\sqrt[4x^2]=4x` `⇔x^2 / (y-1)≥4x-4(y-1) (1)` `y^2 / (x-1)+4(x-1)≥2\sqrt[4y^2]=4y` `⇔y^2 / (x-1)≥4y-4(x-1) (2)` từ` (1);(2)` `⇒B≥4x-4(y-1)+4y-4(x-1)` `⇔B≥8` `”=” xẩy ra khi ` `x=y=2` Bình luận
`A=(x-2019)^2+(x+2020)^2`
`⇔A=x^2-4038x+2019^2+x^2+4040x+2020^2`
`⇔A=2x^2+2019^2+2x+2020^2`
`⇔A=2x^2+2019^2+2x+2020^2`
`⇔A=2(x^2+x+1/4)+2019^2-1/2+2020^2`
`⇔A=2(x+1/2)^2+2019^2-1/2+2020^2≥2019^2-1/2+2020^2`
`”= xẩy ra khi”`
`x+1/2=0`
`⇒x=-1/2`
`⇒minA=2019^2-1/2+2020^2 ` khi` x=-1/2`
`B=x^2 / (y-1) + y^2/(x-1)`
theo `co-si`
`x^2 / (y-1)+4(y-1)≥2\sqrt[4x^2]=4x`
`⇔x^2 / (y-1)≥4x-4(y-1) (1)`
`y^2 / (x-1)+4(x-1)≥2\sqrt[4y^2]=4y`
`⇔y^2 / (x-1)≥4y-4(x-1) (2)`
từ` (1);(2)`
`⇒B≥4x-4(y-1)+4y-4(x-1)`
`⇔B≥8`
`”=” xẩy ra khi `
`x=y=2`
Mình chỉ biết lam cau b thoi mong bạn thông cảm