Ta có: x^2+2y^2-2xy+2x+2-4y=0

=> x^2 -2xy+y^2+ 2x-2y+1+y^2-2y+1=0

=> (x-y)^2+ 2(x-y)+1 + (y-1)^2=0

=> (x-y+1)^2+(y-1)^2=0

mà (x-y+1)^2> hoặc=0 với mọi x;y

(y-1)^2> hoặc=0 với mọi x;y

nên x-y+1=0;y-1=0

=> y=1; x=0

A \(=\) x\(^2\) +2y\(^2\) - 2xy- 4y + 5

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

\(=\) ( x + y )\(^2\) + ( y - 2 )\(^2\) + 1

Vì ( x + y )\(^2\) và ( y - 2 )\(^2\) > 0 ∀ x và y

Nên ( x + y )\(^2\) + ( y - 2 )\(^2\) + 1 > 1 ∀ x và y

Vậy A có giá trị nhỏ nhất là 1 khi

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\text{x + y =0}\\y-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)

B = 5x\(^2\) + 8xy + 5y\(^2\) - 2x = 2y ???

Đề bài câu B sai

A= (4x²+8xy+4y²)+ (x²-2x+1)-1+(y²+2y+1)-1+2019= 4(x+y)² + (x-1)²+(y+1)²+2017 \(\ge\)2017

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

Vậy MinA= 2017 khi x=1; y=-1

A=5+ (-x²+2x) +(-4y²-4y)= -(x²-2x+1)+1-(4y²+4y+1)+1+5=-(x-1)²-(2y+1)² +7 \(\le\)7

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

Vậy Max A bằng 7 khi x=1; y=-1/2

a) \(\Leftrightarrow x^2-2xy+y^2+y^2-4y+4+1\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-2\right)^2+1\ge1\)

 vậy Min=1 khi x-y =0 , y-2 = 0 <=> x=y,y=2=>x=y=2

|Mấy câu sau tương tự nếu ko biết thì nói nha

,

a)\(x^2+5y^2-2xy+4y+1=0\)

\(x^2+2xy+y^2+4y^2+4y+1=0\)

\(\left(x+y\right)^2+\left(2y+1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}x+y=0\\2y+1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-y\\y=-\frac{1}{2}\left(1\right)\end{cases}}\)

      Từ (1) ta đc: x = 1/2

b)\(5x^2+5y^2+8xy-2x+2y+2=0\)

\(4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0\)

\(\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}2x+2y=0\\x-1=0\\y+1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-y\\x=1\\y=-1\end{cases}}\)

CÂU B Sao bạn làm được vậy

Bài làm:

a) \(x^2+5y^2-2xy+4y+1=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(2y+1\right)^2=0\)

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(2y+1\right)^2\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{cases}\Rightarrow}x=y=-\frac{1}{2}\)

Vậy \(x=y=-\frac{1}{2}\)

b) \(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(x+1\right)^2=0\)

Vì \(\hept{\begin{cases}4\left(x+y\right)^2\ge0\\\left(x-1\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}4\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}}\)