\(C=3x^2-5x-8=3\left(x^2-\frac{5}{3}x+\frac{25}{36}\right)-\frac{121}{12}=3\left(x-\frac{5}{6}\right)^2-\frac{121}{12}\ge-\frac{121}{12}\)

Dấu "=" xảy ra khi x=5/6

Vậy Cmin=-121/12 khi x=5/6

\(C=3x^2-5x-8=3\left(x^2-2x.\frac{5}{6}+\frac{25}{36}\right)-\frac{25}{12}-8\)

\(=3\left(x-\frac{5}{6}\right)^2-\frac{121}{12}\)\(\Rightarrow C\ge-\frac{121}{12}\)(Do \(3\left(x-\frac{5}{6}\right)^2\ge0\))

Vậy \(Min_C=-\frac{121}{12}\). Dấu "=" xảy ra <=> \(x=\frac{5}{6}.\)

F = 5x² + 2y² + 4xy - 2x + 4y + 8

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

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

\(\hept{\begin{cases}\left(2x+y\right)^2\\\left(x-1\right)^2\\\left(y+2\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\forall x,y\)

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

Vậy MinF = 3 <=> x = 1 , y = -2

G = 5x² + 5y² + 8xy + 2y + 2020

= x² + ( 4x² + 8xy + 4y² ) + ( y² + 2y + 1 ) + 2019

= x² + ( 2x + 2y )² + ( y + 1 )² + 2019

\(\hept{\begin{cases}x^2\\\left(2x+2y\right)^2\\\left(y+1\right)^2\end{cases}}\ge0\forall x,y\Rightarrow x^2+\left(2x+2y\right)^2+\left(y+1\right)^2+2019\ge2019\forall x,y\)

Tuy nhiên đẳng thức không xảy ra :P

a)x²+ 5x+8

\(=\left(x^2+5x+\frac{25}{4}+\frac{7}{4}\right)\)

\(=\left(x^2+5x+\left(\frac{5}{2}\right)^2\right)+\frac{7}{4}\) 

\(=\left(x+\frac{5}{2}\right)^2+\frac{7}{4}\ge0+\frac{7}{4}=\frac{7}{4}\)

Dấu = khi \(x=-\frac{5}{2}\)

Vậy \(Min_A=\frac{7}{4}\Leftrightarrow x=-\frac{5}{2}\)

\(x^2+5x+8\)

\(=x^2+2x\frac{5}{2}+\frac{25}{4}-\frac{25}{4}+8\)

\(=\left(x+\frac{5}{2}\right)^2+\frac{7}{4}\) 

Vì \(\left(x+\frac{5}{2}\right)^2\ge0\)

Nên \(\left(x+\frac{5}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)

Vậy GTNN của đa thức là: \(\frac{7}{4}\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

khó thế

Ta có: A = 2x² - 5x + 3 = 2(x² - 5/2x + 25/16) - 1/8 = 2(x - 5/4)² - 1/8 \(\le\)-1/8 \(\forall\)x

Dấu "=" xảy ra <=> x - 5/4 = 0 <=> x = 5/4

Vậy MinA = -1/8 <=> x = 5/4

\(A=2x^2-5x+3=2\left(x^2-\frac{5}{2}x+\frac{3}{2}\right)\)

\(=2\left(x^2-\frac{5}{2}x+\frac{25}{16}-\frac{1}{16}\right)\)

\(=2\left[\left(x-\frac{5}{4}\right)^2-\frac{1}{16}\right]\)

\(=2\left[\left(x-\frac{5}{4}\right)^2\right]-\frac{1}{8}\ge\frac{-1}{8}\)

\(2x^2-5x+3\)\(=2\left(x^2-\frac{5}{2}x+\frac{3}{2}\right)\)

                                  \(=2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{1}{16}\right)\)

                                  \(=2\left(x-\frac{5}{4}\right)^2-\frac{1}{8}\)

Vì \(2\left(x-\frac{5}{4}\right)^2\ge0\forall x\)=>\(2\left(x-\frac{5}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\forall x\)

MIN A = \(-\frac{1}{8}\Leftrightarrow x=\frac{5}{4}\)

\(P=5x^2+y^2-2x(y+8)+2023\\=5x^2+y^2-2xy-16x+2023\\=(x^2-2xy+y^2)+(4x^2-16x+16)+2007\\=(x-y)^2+4(x^2-4x+4)+2007\\=(x-y)^2+4(x-2)^2+2007\)

Ta thấy: \(\left(x-y\right)^2\ge0\forall x;y\)

              \(4\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow (x-y)^2+4(x-2)^2\ge0\forall x;y\\\Rightarrow P=(x-y)^2+4(x-2)^2+2007\ge2007\forall x;y\)

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

Vậy \(Min_P=2007\) khi \(x=y=2\).

\(\text{#}Toru\)

\(P=5x^2+y^2-2x\left(y+8\right)+2023\)

\(=x^2-2xy+y^2+4x^2-16x+2023\)

\(=\left(x-y\right)^2+4x^2-16x+16+2007\)

\(=\left(x-y\right)^2+\left(2x-4\right)^2+2007>=2007\)

Dấu = xảy ra khi x-y=0 và 2x-4=0

=>x=y=2

P = 5x² + y² - 2x(y + 8) + 2023

= 4x² + x² + y² - 2xy - 16x + 16 + 2007

= (x² - 2xy + y²) + (4x² - 16x + 16) + 2007

= (x - y)² + (2x - 4)² + 2007

Do (x - y)² ≥ 0 với mọi x, y R

(2x - 4)² ≥ 0 với mọi x R

⇒ (x - y)² + (2x - 4)² ≥ 0 với mọi x, y ∈ R

⇒ (x - y)² + (2x - 4)² + 2007 ≥ 2007 với mọi x, y ∈ R

Vậy GTNN của P là 2007 khi x = y = 2