Câu 1 :

\(E=4x^2+y^2-4x-2y+3\)

\(E=\left(2x\right)^2-2\cdot2x\cdot1+1^2+y^2-2\cdot y\cdot1+1^2+1\)

\(E=\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\forall x;y\)

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

Câu 2 :

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

\(G=x^2+2xy+y^2+y^2-2.y\cdot1+1^2-1\)

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

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

Còn câu F bạn ơi. Giúp Gk vs

\(F=\frac{3}{2x^2+x+1}=\frac{3}{2\left(x^2+\frac{x}{2}+\frac{1}{2}\right)}=\frac{3}{2\left(x^2+2x\cdot\frac{1}{4}+\frac{1}{16}\right)+\frac{7}{8}}=\frac{3}{2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}}\)

Vi \(2\left(x+\frac{1}{4}\right)^2\ge0\Rightarrow2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}\ge8\)

\(\Rightarrow\frac{1}{2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}}\le\frac{1}{\frac{7}{8}}\Rightarrow F=\frac{3}{2\left(x+\frac{1}{4}\right)^2+\frac{7}{8}}\le\frac{3}{\frac{7}{8}}=\frac{24}{7}\)

Dấu "=" xảy ra <=>x+1/4=0<=>x=-1/4

\(S=\left(x^2+y^2+1+2xy+2x+2y\right)+\left(y^2-4y+4\right)+2021\)

\(S=\left(x+y+1\right)^2+\left(y-2\right)^2+2021\ge2021\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(-3;2\right)\)

x^2+y^2+2x+2y+2xy+5

bài này rễ quá

giá trị nhỏ nhất của A là 5 nhé

a) \(A=4x^2-12x+100=\left(2x\right)^2-12x+3^2+91=\left(2x-3\right)^2+91\)

Ta có: \(\left(2x-3\right)^2\ge0\forall x\inℤ\)

\(\Rightarrow\left(2x-3\right)^2+91\ge91\)

hay A \(\ge91\)

Dấu "=" xảy ra <=> \(\left(2x-3\right)^2=0\)

<=> 2x-3=0

<=> 2x=3

<=> \(x=\frac{3}{2}\)

Vậy Min A=91 đạt được khi \(x=\frac{3}{2}\)

b) \(B=-x^2-x+1=-\left(x^2+x-1\right)=-\left(x^2+x+\frac{1}{4}-\frac{5}{4}\right)=-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\)

Ta có: \(-\left(x+\frac{1}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\) hay B\(\le\frac{5}{4}\)

Dấu "=" \(\Leftrightarrow-\left(x+\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy Max B=\(\frac{5}{4}\)đạt được khi \(x=\frac{-1}{2}\)

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

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

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

Ta có: 

\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x;y\inℤ\\x^2\ge0\forall x\inℤ\end{cases}}\)

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

hay C\(\ge\)1

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

Vậy Min C=1 đạt được khi y=1 và x=0

\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+x^2+6x+9+1978\)

\(=\left(x-y\right)^2+2\left(x-y\right)+1+\left(x+3\right)^2+1978\)

\(=\left(x-y+1\right)^2+\left(x+3\right)^2+1978\ge1978\)

\(A_{min}=1978\) khi \(\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

Ta có:

A = 2x² + 2xy + y² - 2x + 2y + 2

A = (x² + 2xy + y²) + 2(x + y) + 1 + (x² - 4x + 4) - 3

A = (x + y)² + 2(x + y) + 1 + (x - 2)² - 3

A = (x + y + 1)² + (x - 2)² - 3 \(\ge\)-3 \(\forall\)x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y+1=0\\x-2=0\end{cases}}\) <=> \(\hept{\begin{cases}y=-x-1\\x=2\end{cases}}\) <=> \(\hept{\begin{cases}y=-2-1=-3\\x=2\end{cases}}\)

Vậy MinA = -3 <=> x = 2 và y = -3

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

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

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

Ta thấy \(\left(x+y+1\right)^2\ge0\)     \(\forall x,y\)

             \(\left(x-2\right)^2\ge0\)            \(\forall x\)

=> \(\left(x+y+1\right)^2+\left(x-2\right)^2\ge0\)     \(\forall x,y\)

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

hay \(A\ge-1\)

\(MinA=-1\)\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\x-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

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

\(=x^2-4x+4+x^2+y^2+1+2x+2y+2xy-3\)

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

Dấu \(=\)khi \(\hept{\begin{cases}x-2=0\\x+y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=-3\end{cases}}\).

Lời giải:
Ta có:

$P=2x^2+y^2+2xy+5x+y+\frac{37}{4}$

$=(x^2+y^2+2xy)+x^2+5x+y+\frac{37}{4}$

$=(x+y)^2+(x+y)+(x^2+4x)+\frac{37}{4}$

$=(x+y)^2+(x+y)+\frac{1}{4}+(x^2+4x+4)+5$

$=(x+y+\frac{1}{2})^2+(x+2)^2+5\geq 5$

Vậy $P_{\min}=5$. Giá trị này đạt tại:

$x+y+\frac{1}{2}=x+2=0$

$\Leftrightarrow x=-2; y=\frac{3}{2}$

Lời giải:
Ta có:

$P=2x^2+y^2+2xy+5x+y+\frac{37}{4}$

$=(x^2+y^2+2xy)+x^2+5x+y+\frac{37}{4}$

$=(x+y)^2+(x+y)+(x^2+4x)+\frac{37}{4}$

$=(x+y)^2+(x+y)+\frac{1}{4}+(x^2+4x+4)+5$

$=(x+y+\frac{1}{2})^2+(x+2)^2+5\geq 5$

Vậy $P_{\min}=5$. Giá trị này đạt tại:

$x+y+\frac{1}{2}=x+2=0$

$\Leftrightarrow x=-2; y=\frac{3}{2}$