\(M=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+6=\left(x-y\right)^2+\left(x-2\right)^2+6\)

\(\min M=6\) với \(x=y=2\)

\(M=2x^2+y^2-2xy-4x-10\)

\(=x^2-2xy+y^2+x^2-4x+4-14\)

\(=\left(x-y\right)^2+\left(x-2\right)^2-14\ge-14\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-y=0\\ x-2=0\end{cases}\Rightarrow x=y=2\)

\(M=\left(x-y\right)^2-2\left(x-y\right)+1+9=\left(x-y-1\right)^2+9\ge9\)

\(\min M=9\)

\(M=x^2+y^2-2xy-2x+2y+10\)

\(=\left(x-y\right)^2-2\left(x-y\right)+1+9\)

\(=\left(x-y-1\right)^2+9\ge9\forall x,y\)

Dấu '=' xảy ra khi x-y-1=0

=>x=y+1

a) \(2x^2+y^2+4x-2y-2xy+10\)

\(=x^2+x^2+y^2+4x-2y-2xy+4+6\)

\(=\left(x^2-2xy+y^2\right)+\left(x^2+4x+4\right)-2\left(y-3\right)\)

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

.......................chắc không phải cách làm này đâu!

b) \(5x^2+y^2+2xy-4x\)

\(=x^2+4x^2+y^2+2xy-4x\)

\(=\left(x^2+2xy+y^2\right)+x^2-4x\)

\(\left(x+y\right)^2+x^2-4x\)

a, \(2x^2\)+\(y^2\)+\(4x-2y-2xy+10\)\(=y^2\)\(-x^2\)\(-1+2x-2y-2xy+3x^2+2x+11\)\(=\left(y-x-1^{ }\right)^2\)\(+3\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+\frac{32}{3}\)\(=\left(y-x-1\right)^2+3\left(x+\frac{1}{3}\right)^2+\frac{32}{3}\)\(\ge\frac{32}{3}\)

VẬY GTNN CỦA BIỂU THỨC \(=\frac{32}{3}\)KHI \(y-x-1=0;x+\frac{1}{3}=0\Rightarrow x=\frac{-1}{3};y=\frac{2}{3}\)

\(2x^2+y^2+2xy-6x-2y+10\)

\(=\left(x^2-4x+4\right)+\left(x^2+y^2+1+2xy-2y-2x\right)+5\)

\(=\left(x-2\right)^2+\left(x+y-1\right)^2+5\ge5\)

Đặt \(A=x^2+2y^2+2xy+2x+4y-1\)

\(A=\left(x^2+2xy+y^2\right)+\left(y^2+2y\right)+\left(2x+2y\right)-1\)

\(A=\left[\left(x+y\right)^2+2\left(x+y\right)+1\right]+\left(y^2+2y+1\right)-3\)

\(A=\left(x+y+1\right)^2+\left(y+1\right)^2-3\ge-3\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}}\)

Vậy GTNN của \(A\) là \(-3\) khi \(x=0\) và \(y=-1\)

Chúc bạn học tốt ~ 

Đặt \(B=-x^2-2x-y^2-8y-10\)

\(-B=\left(x^2+2x+1\right)+\left(y^2+8y+16\right)-7\)

\(-B=\left(x+1\right)^2+\left(y+4\right)^2-17\ge-17\)

\(B=-\left(x+1\right)^2-\left(y+4\right)^2+17\le17\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}-\left(x+1\right)^2=0\\-\left(y+4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-4\end{cases}}}\)

Vậy GTLN của \(B\) là \(17\) khi \(x=-1\) và \(y=-4\)

Chúc bạn học tốt ~ 

a)\(2x^2+y^2+4x-2y-2xy+10=2x^2+y^2+4x-2y\left(x+1\right)+10\)

\(=y^2-2y\left(x+1\right)+2\left(x^2+2x+1\right)+8\)

\(=y^2-2y\left(x+1\right)+2\left(x+1\right)^2+8\)

\(=\left(y+x+1\right)^2+\left(x+1\right)^2+8\ge8\)

Dấu "=" xảy ra khi x=-1 và y=0

b)\(5x^2+y^2+2xy-4x=\left(x^2+2xy+y^2\right)+\left(4x^2-4x+1\right)-1\)

\(=\left(x+y\right)^2+\left(2x-1\right)^2-1\ge-1\)

Dấu "=" xảy ra khi x=1/2 và y=-1/2

\(P=2x^2+y^2+2xy-6x-2y+10\)

\(P=\left(x^2+y^2+1^2-2y-2x\right)+\left(x^2-4x+4\right)+5\)

\(P=\left(x+y-1\right)^2+\left(x-2\right)^2+5\)

\(\left\{{}\begin{matrix}\left(x+y-1\right)^2\ge0\\\left(x-2\right)^2\ge0\end{matrix}\right.\) \(\Rightarrow P\ge5\) đẳng thức khi \(\left\{{}\begin{matrix}x-2=0\\x+y-1=0\end{matrix}\right.\) => x=2 và y=-1

2x² + y² + 2xy - 6x - 2y + 10

= x² + y² + 1² + 2xy - 2x - 2y + x² - 4x + 4 + 5

= (x + y - 1)² + (x - 2)² + 5 \(\ge\) 5

Dấu ''='' xảy ra khi \(\left\{{}\begin{matrix}x+y-1=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=2\end{matrix}\right.\)

Vậy Min = 5 khi x = 2 và y = - 1

\(A=2x^2+2xy+y^2-2x+2y+2\)

\(\Rightarrow2A=4x^2+4xy+2y^2-4x+4y+4\)

              \(=\left(4x^2+4xy+y^2\right)-2\left(2x+y\right).1+1+y^2+6y+9-6\)

               \(=\left(2x+y\right)^2-2\left(2x+y\right)+1+\left(y+3\right)^2-6\)

                \(=\left(2x+y-1\right)^2+\left(y+3\right)^2-6\)

vì \(\left(2x+y-1\right)^2\ge0\forall x,y;\left(y+3\right)^2\ge0\forall y\)nên

\(2A=\left(2x+y-1\right)+\left(y+3\right)-6\ge-6\forall x,y\)

hay \(2A\ge-6\Rightarrow A\ge-3\Rightarrow minA=-3\Leftrightarrow\hept{\begin{cases}2x+y-1=0\\y+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)