\(P=x^2-xy+y^2-3x-3y+16\)

\(2P=2x^2-2xy+2y^2-6x-6y+32\)

\(2P=\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)+\left(y^2-6x+9\right)+14\)

\(2P=\left(x-y\right)^2+\left(x-3\right)^2+\left(y-3\right)^2+14\ge14\)

Dấu "=" xảy ra tại \(x=y=3\)

Mình đoán đề bị sai,mình đã sửa rồi nhé !

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Đừng khóa nick nha olm

\(M=x^2+y^2+xy-3x-3y+2018\)

\(=x^2+2x\frac{\left(y-3\right)}{2}+\left(\frac{y-3}{2}\right)^2+y^2-3y+2018-\left(\frac{y-3}{2}\right)^2\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3y^2-6y+8063}{4}\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3\left(y^2-2y+1\right)}{4}+2015\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3\left(y-1\right)^2}{4}+2015\ge2015\)

\("="\Leftrightarrow x=y=1\)

Cảm ơn bạn nhiều nha

\(A=\frac{1}{4}\left(4x^2+4y^2+4xy-12x-12y\right)+2006\)

\(A=\frac{1}{4}\left(x^2+4y^2+9+4xy-6x-12y\right)+\frac{3}{4}\left(x^2-2x+1\right)+2003\)

\(A=\frac{1}{4}\left(x+2y-3\right)^2+\frac{3}{4}\left(x-1\right)^2+2003\ge2003\)

\(\Rightarrow A_{min}=2003\) khi \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

\(x^2+y^2+xy+3x-3y+9=0\)

\(\Leftrightarrow2x^2+2y^2+2xy+6x-6y+18=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+\left(y^2-6y+9\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+3\right)^2+\left(y-3\right)^2=0\)

\(\Leftrightarrow x=-3;y=3\)

Thay vào:\(Q=\left(3-3+1\right)^{2017}+\left(2-3\right)^{2018}=2\)

Bài 1 dùng tam thức bậc 2, bài 2 chia cả tử và mẫu cho y², đặt x/y=t rồi làm tương tự bài 1

Đặt \(\dfrac{x}{y}=t\)

\(Q=\dfrac{\dfrac{x^2-xy+2y^2}{y^2}}{\dfrac{x^2-xy+y^2}{y^2}}=\dfrac{\dfrac{x^2}{y^2}-\dfrac{x}{y}+2}{\dfrac{x^2}{y^2}-\dfrac{x}{y}+1}=\dfrac{t^2-t+2}{t^2-t+1}\)

\(\Rightarrow Qt^2-Qt+Q=t^2-t+2\Leftrightarrow t^2\left(Q-1\right)-t\left(Q-1\right)+Q-2=0\)

\(\Delta=\left(Q-1\right)^2-4\left(Q-1\right)\left(Q-2\right)\ge0\)

\(\Rightarrow1\le Q\le\dfrac{7}{3}\)

\(P=x^2+3x+y^2+3y+\frac{9}{x^2+y^2+1}\)

\(=x^2+y^2+1+\frac{9}{x^2+y^2+1}+3x+3y-1\)

\(\ge2.3.\frac{\sqrt{x^2+y^2+1}}{\sqrt{x^2+y^2+1}}+2.3.\sqrt{xy}-1\)

\(=6+6-1=11\)

Dấu = xảy ra khi x = y = 1

Ta có

2B = 2x² + 2y² + 2xy - 6x - 6y + 4026

= (x² + 2xy + y²) - (4x + 4y) + (x² - 2x + 1) + (y² - 2y + 1) + 4 + 4020

= (x + y)² - 4(x + y) + 4 + (x - 1)² + (y - 1)² + 4020

= (x + y -2)² + (x - 1)² + (y - 1)² + 4020 \(\ge4020\)

=> B\(\ge2010\)

Đạt được khi x = y = 1