A= X^2- 6X +9 + y^2 -22y + 121+ z^2+12z+ 36+2019

= (x-3)²+(y-11)²+(z+6)²+2019

Lại có (x-3)²+(y-11)²+(z+6)²\(\ge\)0

=> A\(\ge\)2019

Vậy Min A = 2019 <=> x= 3; y=11; z= -6

\(x^2+y^2+z^2-6x-22y+12z+166=0\)

\(\Leftrightarrow x^2+y^2+z^2-6x-22y+12z+121+9+36=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(y^2-22y+121\right)+\left(z^2+12z+36\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y-11\right)^2+\left(z+6\right)^2=0\)

\(\hept{\begin{cases}\left(x-3\right)^2\ge0\\\left(y-11\right)^2\ge0\\\left(z+6\right)^2\ge0\end{cases}}\Rightarrow\hept{\begin{cases}\left(x-3\right)^2=0\\\left(y-11\right)^2=0\\\left(z+6\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-3=0\\y-11=0\\z+6=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=11\\z=-6\end{cases}}\)

\(M=2x^2+9y^2-6xy-6x-12y+2028\\ =3\left(x^2-2xy+y^2\right)-\left(x^2+6x+9\right)+6\left(y^2-2y+1\right)+2025\\ =\left(x-y\right)^2-\left(x-3\right)^2+6\left(y-1\right)^2+2025\ge2025\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=1\end{matrix}\right.\) (vô lí) nên dấu \("="\) ko thể xảy ra

\(N=x^2-4xy+5y^2+10x-22y+28\\ =\left(x^2+4y^2+25-4xy-20y+10x\right)+\left(y^2-2y+1\right)+2\\=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-2y=5\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

\(M=2x^2+9y^2-6xy-6x-12y+2028=\left(x+2\right)^2-6y\left(x+2\right)+9y^2+\left(x-5\right)^2+1999=\left(x+2-3y\right)^2+\left(x-5\right)^2+2019\ge1999\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\dfrac{7}{3}\end{matrix}\right.\)

\(N=x^2-4xy+5y^2+10x-22y+28=\left(x+5\right)^2-4y\left(x+5\right)+4y^2+\left(y-1\right)^2+2=\left(x+5-2y\right)^2+\left(y-1\right)^2+2\ge2\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

\(A=x^2-6x+11\)

\(A=\left(x^2-6x+9\right)+2\)

\(A=\left(x-3\right)^2+2\ge2\)

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

\(\Leftrightarrow\)\(x-3=0\)

\(\Leftrightarrow\)\(x=3\)

Vậy GTNN của \(A\) là \(2\) khi \(x=3\)

\(B=x^2-20x+101\)

\(B=\left(x^2-20x+100\right)+1\)

\(B=\left(x-10\right)^2+1\ge1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left(x-10\right)^2=0\)

\(\Leftrightarrow\)\(x-10=0\)

\(\Leftrightarrow\)\(x=10\)

Vậy GTNN của \(B\) là \(1\) khi \(x=10\)

Chúc bạn học tốt ~ 

\(A=x^2-6x+11\)

\(A=\left(x^2-6x+9\right)+2\)

\(A=\left(x-3\right)^2+2\)

Mà  \(\left(x-3\right)^2\ge0\)

\(\Rightarrow A\ge2\)

Dấu "=" xảy ra khi :  \(x-3=0\Leftrightarrow x=3\)

Vậy  \(A_{Min}=2\Leftrightarrow x=3\)

b) \(B=x^2-20x+101\)

\(B=\left(x^2-20x+100\right)+1\)

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

Mà  \(\left(x-10\right)^2\ge0\)

\(\Rightarrow B\ge1\)

Dấu "=" xảy ra khi :  \(x-10=0\Leftrightarrow x=10\)

Vậy  \(B_{Min}=1\Leftrightarrow x=10\)

c)  \(C=x^2-4xy+5y^2+10x-22y+28\)

\(C=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+28\)

\(C=\left[\left(x-2y\right)^2+2\left(x-2y\right).5+25\right]+\)\(\left(y^2-2y+1\right)+2\)

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

Mà  \(\left(x-2y+5\right)^2\ge0\)

      \(\left(y-1\right)^2\ge0\)

\(\Rightarrow C\ge2\)

Dấu "=" xảy ra khi : 

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vây  \(C_{Min}=2\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)

\(D=4x-x^2+3\)

\(-D=x^2-4x-3\)

\(-D=\left(x^2-4x+4\right)-7\)

\(-D=\left(x-2\right)^2-7\)

Mà  \(\left(x-2\right)^2\ge0\)

\(\Rightarrow-D\ge-7\)

\(\Leftrightarrow D\le7\)

Dấu "=" xảy ra khi :  \(x-2=0\Leftrightarrow x=2\)

Vậy  \(D_{Max}=7\Leftrightarrow x=2\)

\(E=-x^2+6x-11\)

\(-E=x^2-6x+11\)

\(-E=\left(x^2-6x+9\right)+2\)

\(-E=\left(x-3\right)^2+2\)

Mà  \(\left(x-3\right)^2\ge0\)

\(\Rightarrow-E\ge2\)

\(\Leftrightarrow E\le-2\)

Dấu "=" xảy ra khi  \(x-3=0\Leftrightarrow x=3\)

Vậy  \(E_{Max}=-2\Leftrightarrow x=3\)

a) x² - 6x + 11 = ( x² - 6x + 9 ) + 2 = ( x - 3 )² + 2 ≥ 2 ∀ x

Dấu "=" xảy ra khi x = 3

=> GTNN của bthuc = 2 <=> x = 3

b) x² - 20x + 101 = ( x² - 20x + 100 ) + 1 = ( x - 10 )² + 1 ≥ 1 ∀ x

Dấu "=" xảy ra khi x = 10

=> GTNN của bthuc = 1 <=> x = 10

c) x² - 4xy + 5y² + 10x - 22y + 28

= ( x² - 4xy + 4y² + 10x - 20y + 25 ) + ( y² - 2y + 1 ) + 2

= [ ( x² - 4xy + 4y² ) + ( 10x - 20y ) + 25 ] + ( y - 1 )² + 2

= [ ( x - 2y )² + 2( x - 2y ).5 + 5² ] + ( y - 1 )² + 2

= ( x - 2y + 5 )² + ( y - 1 )² + 2 ≥ 2 ∀ x, y

Dấu "=" xảy ra khi x = -3 ; y = 1

=> GTNN của bthuc = 2 <=> x = -3 ; y = 1

a) \(x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\)

ta có: \(\left(x-3\right)^2\ge0\forall x\)=> \(\left(x-3\right)^2+2\ge2\)

dấu "=" xảy ra khi \(x-3=0\Leftrightarrow x=3\)

Vậy biểu thức đạt GTNN là 2 khi chỉ khi x = 3

       \(a,x^2-6x+11=x^2-2.x.3+9+2=\left(x-3\right)^2+2\)

Mà \(\left(x-3\right)^2\ge0\)\(\Rightarrow\left(x-3\right)^2+2\ge2\)

       \(\Rightarrow GTNN\)của \(x^2-6x+11\) là \(2\)

Dấu "=" khi và chỉ khi x=3