\(F=2x^2+2y^2-x-y+11\)

\(F=2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+2\left(y^2-\frac{1}{2}y+\frac{1}{16}\right)+\frac{87}{8}\)

\(F=2\left(x-\frac{1}{4}\right)^2+2\left(y-\frac{1}{4}\right)^2+\frac{87}{8}\ge\frac{87}{8}\)

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

Vậy GTNN của \(F\) là \(\frac{87}{8}\) khi \(x=y=\frac{1}{4}\)

Chúc bạn học tốt ~ 

B = 2\(x^2\) - 4\(x\) - 8

B = 2(\(x^2\) - 2\(x\) + 4)  - 16

B = 2(\(x-2\))² - 16 

Vì (\(x-2\))² ≥ 0 ∀ \(x\) ⇒ 2(\(x-2\))² ≥ 0 ∀ \(x\)

⇒ 2(\(x-2\))²  - 16 ≥ -16 ∀ \(x\)

Dấu bằng xảy ra khi  (\(x-2\))² = 0 ⇒ \(x-2=0\) ⇒ \(x=2\)

Vậy B_min = -16 khi \(x=2\)

Tìm min của C biết:

C = \(x^2\) - 2\(xy\) + 2y² + 2\(x\) - 10y + 17

C = (\(x^2\) - 2\(xy\) + y²) + 2(\(x\) - y) + y² - 8y + 16 + 1

C = (\(x\) - y)² + 2(\(x\) - y) + 1  + (y² - 8y + 16) 

C = (\(x-y+1\))² + (y - 4)² 

Vì (\(x\) - y + 1)² ≥ 0 ∀ \(x;y\); (y - 4)² ≥ 0 ∀ y

Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x-y+1=0\\y=4\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x-4+1=0\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-1+4\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

Vậy C_min = 0 khi (\(x;y\)) = (3; 4)

D = \(x^2\) - \(xy\) + y² - 2\(x\) - 2y

D=[\(x^2\)-2\(x\)\(\dfrac{y}{2}\)+(\(\dfrac{y}{2}\))²]-(2\(x\)-2\(\dfrac{y}{2}\)) +1 +(\(\dfrac{3}{4}\)y²-2.\(\dfrac{\sqrt{3}}{2}\)y .\(\sqrt{3}\) +3) - 4

D = (\(x-\dfrac{y}{2}\))² - 2(\(x-\dfrac{y}{2}\))+ 1 + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4

D = (\(x-\dfrac{y}{2}\) - 1)² + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4

Vì (\(x-\dfrac{y}{2}\) - 1)² ≥  0 ∀ \(x\);y và (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² ≥ 0 ∀ y 

Vậy (\(x-\dfrac{y}{2}\) - 1)² + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4 ≥ - 4 ∀ \(x;y\)

Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\\dfrac{\sqrt{3}}{2}y-\sqrt{3}=0\end{matrix}\right.\)

      ⇒ \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\\sqrt{3}.\left(\dfrac{1}{2}y-1\right)=0\end{matrix}\right.\)

  ⇒ \(\left\{{}\begin{matrix}x=1+\dfrac{1}{2}y\\\dfrac{1}{2}y=1\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=1+1\\y=1:\dfrac{1}{2}\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Vậy D_min = - 4 khi (\(x;y\)) =(2; 2)

\(B=2x^2-4x-8=2\left(x^2-2x-4\right)\)

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

\(=2\left[\left(x-1\right)^2-5\right]\)

\(=2\left(x-1\right)^2-10\ge-10\)

Vậy \(B_{min}=-10\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(F=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)

Đặt \(x^2+5x+4=t\)

\(\RightarrowĐT=t\left(t+2\right)=t^2+2t+1-1\)

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

hay \(\left(x^2+5x+5\right)^2-1\ge-1\)

Vậy \(F_{min}=-1\Leftrightarrow x^2+5x+5=0\)

\(\Leftrightarrow x^2+5x+\frac{25}{4}-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=\frac{5}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{2}=\sqrt{\frac{5}{4}}\\x+\frac{5}{2}=-\sqrt{\frac{5}{4}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{5}{4}}-\frac{5}{2}\\x=-\sqrt{\frac{5}{4}}-\frac{5}{2}\end{cases}}\)

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

\(=-\left[\left(x-2\right)^2-4\right]=-\left(x-2\right)^2+4\le4\)

Vậy \(G_{max}=4\Leftrightarrow x-2=0\Leftrightarrow x=2\)

\(H=25-x-5x^2=-5\left(x^2+\frac{x}{5}-5\right)\)

\(=-5\left(x^2+2x.\frac{1}{10}+\frac{1}{100}-\frac{501}{100}\right)\)

\(=-5\left[\left(x+\frac{1}{10}\right)^2-\frac{501}{100}\right]\)

\(=-5\left(x+\frac{1}{10}\right)^2+\frac{101}{20}\le\frac{101}{2}\)

Vậy \(H_{max}=\frac{101}{2}\Leftrightarrow x+\frac{1}{10}=0\Leftrightarrow x=-\frac{1}{10}\)

Lỗi Latex

\(2x^2+y^2-xy+x-2y=\dfrac{1}{4}x^2-x\left(y-1\right)+\left(y-1\right)^2+\dfrac{7}{4}x^2-1=\left(\dfrac{1}{2}x-y+1\right)^2+\dfrac{7}{4}x^2-1\ge-1\)

\(min=-1\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

Ta có:

\(A=\sqrt{\left(x-3\right)^2+2\left(y+1\right)^2}+\sqrt{\left(x+1\right)^2+3\left(y+1\right)^2}\)

Áp dụng bđt Minkowski, ta có:

\(\Rightarrow A=\sqrt{\left(x-3\right)^2+2\left(y+1\right)^2}+\sqrt{\left(x+1\right)^2+3\left(y+1\right)^2}\)

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

\(A=\sqrt{4^2+\left(\sqrt{2}+\sqrt{3}\right)^2\left(y+1\right)^2}\ge\sqrt{4^2}=4\)

\(\Rightarrow A\ge4.Đ\text{TXR}\Leftrightarrow\orbr{\begin{cases}x=1;y=-1\\x=3;y=-1\end{cases}}\)

Dấu "=" xảy ra khi (x; y) = (3; -1)