D ez nhất :v

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

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

Đẳng thức xảy ra khi x = 1 và y = -2

\(A=\left[\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4\right]+\left(y^2-2y+1\right)+2020\)

\(=\left[\left(x-y\right)^2+2\left(x-y\right).2+2^2\right]+\left(y-1\right)^2+2020\)

\(=\left(x-y+2\right)^2+\left(y-1\right)^2+2020\ge2020\)

Dấu "=" xảy ra khi y = 1 và x - y + 2 = 0 tức là x = y - 2 = -1

\(B=2x^2+y^2-8x+2xy-4y+2025\)

\(=x^2+2xy+y^2-4x-4y+x^2-4x+2025\)

\(=\left(x+y\right)^2-4\left(x+y\right)+4+x^2-4x+4+2017\)

\(=\left(x+y-2\right)^2+\left(x-2\right)^2+2017\ge2017\forall x,y\)

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

A = x² + 2y² + 2xy - 2x - 6y + 6

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

A = (x + y - 1)² + (y - 2)² + 1 \(\ge\)1 \(\forall\)x;y

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

Vậy MinA = 1 khi x = -1 và y = 2

a) \(A=x^2+2y^2+2xy+4x+6y+19\)

\(=\left[\left(x^2+2xy+y^2\right)+2.\left(x+y\right).2+4\right]+\left(y^2+2y+1\right)+14\)

\(=\left[\left(x+y\right)^2+2\left(x+y\right).2+2^2\right]+\left(y+1\right)^2+14\)

\(=\left(x+y+2\right)^2+\left(y+1\right)^2+14\ge14\)

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

b)Đề có gì đó sai sai...

c) Tương tự câu b,em cũng thấy sai sai...HÓng cao nhân giải ạ!

b) \(P=2x^2+y^2+2xy-2y-4\)

\(\Leftrightarrow2P=4x^2+2y^2+4xy-4y-8\)

\(\Leftrightarrow2P=\left(4x^2+4xy+y^2\right)+\left(y^2-4y+4\right)-12\)

\(\Leftrightarrow2P=\left(2x+y\right)^2+\left(y-2\right)^2-12\ge-12\forall x;y\)

Có \(2P\ge-12\Leftrightarrow P\ge-6\)

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

P=\(X^2+2Y^2-2XY+8X+8Y+2017\)

P=\(\dfrac{4X^2+8Y^2-8XY+32Y+32X+8068}{4}\)

P=\(\dfrac{(\sqrt{3}X)^2-2.\sqrt{3}X.\dfrac{4}{\sqrt{3}}Y+\left(\dfrac{4}{\sqrt{3}}Y\right)^2-\left(\dfrac{4}{\sqrt{3}}Y\right)^2+8Y^2+X^2+32X+32Y+8068}{4}\)

P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+\dfrac{8}{3}Y^2+32X+32Y+8068}{4}\)

P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+2.X.16+16^2+(\dfrac{2\sqrt{2}}{\sqrt{3}}Y)^2+2.\dfrac{2\sqrt{2}}{\sqrt{3}}Y.4\sqrt{6}+\left(4\sqrt{6}\right)^2+7716}{4}\)

P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+\left(X+16\right)^2+\left(\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}\right)^2}{4}+1929\ge1929\forall X\in R\)

DẤU = XẢY RA \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y=0\\X+16=0\\\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}=0\end{matrix}\right.\)

A = x² - 2xy + 6y² - 12x + 2y + 45

   = (x² - 2xy + y² - 12x + 12y + 36) + (5y² - 10y + 5) + 4

   = [(x - y)² - 12(x - y) + 6^2] + 5(y² - 2y + 1) + 4

   = (x - y - 6)² + 5(y - 1)² + 4

Vì (x - y - 6)² >= 0 với mọi x, y

   5(y² - 1) >= 0 với mọi y

=> A_min = 4 <=> y = 1, x = 7

tick mk làm cho

\(A=x^2+y^2+2xy+4x+4y+4+y^2+2y+1+14\)

\(A=\left(x+y+2\right)^2+\left(y+1\right)^2+14\ge14\)

\(\Rightarrow A_{min}=14\) khi \(\left\{{}\begin{matrix}y=-1\\x=-1\end{matrix}\right.\)

\(B=2\left(x^2+xy+\frac{y^2}{4}\right)+\frac{1}{2}\left(y^2-4y+4\right)-6\)

\(B=2\left(x+\frac{y}{2}\right)^2+\frac{1}{2}\left(y-2\right)^2-6\ge-6\)

\(\Rightarrow B_{min}=-6\) khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Câu c đề sai, sao vừa có 2xy lại có cả 4xy

Bài 1: Chắc đề là \(a^4-4a^2+4a-1\)

Ta có: \(a^4-4a^2+4a-1=a^4-\left(4a^2-4a+1\right)=a^4-\left(2a-1\right)^2=\left(a^2-2a+1\right)\left(a^2+2a-1\right)=\left(a-1\right)^2\left(a^2+2a-1\right)\)

là số nguyên tố \(\Leftrightarrow\left[{}\begin{matrix}\left(a-1\right)^2=1\\a^2+2a-1=1\end{matrix}\right.\) ( tự giải tiếp)

Bài 2: Làm mẫu một bài thôi nhé

a) Đặt A = \(2x^2+2y^2+2xy-8x-10y+2025\)

\(2A=4x^2+4y^2+4xy-16x-20y+4050\)

\(=\left(2x\right)^2+2.2x\left(y-4\right)+\left(y-4\right)^2-\left(y-4\right)^2+4y^2-20y+4050\)

\(=\left(2x+y-4\right)^2-\left(y^2-8y+16\right)+4y^2-20y+4050\)

\(=\left(2x+y-4\right)^2+3y^2-12y+4034=\left(2x+y-4\right)^2+3\left(y^2-4y+4\right)+4022=\left(2x+y-4\right)^2+3\left(y-2\right)^2+4022\ge4022\forall x,y\)

Vậy min A = 4022 \(\Leftrightarrow\left\{{}\begin{matrix}2x+y-4=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

thanks bạn !

*\(A=x^2+2y^2-2xy-4x-6y-3\)

\(A=x^2-2x\left(y+2\right)+\left(y^2+4y+4\right)+\left(y^2-10y+25\right)-32\)

\(A=x^2-2x\left(y+2\right)+\left(y+2\right)^2+\left(y-5\right)^2-32\)

\(A=\left(x-y-2\right)^2+\left(y-5\right)^2-32\ge-32\)

\(\Rightarrow Min_A=-32\Leftrightarrow x=7;y=5\)

* \(B=4x^2+2y^2-4xy+4x+6y+1\)

\(B=\left(2x\right)^2-\left(4xy+4x\right)+\left(y^2-2y+1\right)+\left(y^2+8y+16\right)-16\)\(B=\left(2x\right)^2-2.2x\left(y-1\right)+\left(y-1\right)^2+\left(y+4\right)^2-16\)\(B=\left(2x-y+1\right)^2+\left(y+4\right)^2-16\ge-16\)

\(\Rightarrow Min_B=-16\Leftrightarrow x=-\dfrac{5}{2};y=-4\)