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

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

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

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

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

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

Có P = x² + 5y² + 4xy + 6x + 16y + 32

         = [(x² + 4xy + 4y²) + 6x + 12y + 9] + (y² + 4y + 2²) + 19

         = [(x + 2y)² + 2(x + 2y).3 + 3² ] + (y + 2)² + 19

         = (x + 2y + 3)² + (y + 2)² + 19

Thấy (x + 2y + 3)² ≥ 0 với mọi x; y

         (y + 2)² ≥ 0 với mọi y

=> (x + 2y + 3)² + (y + 2)² ≥ 0 với mọi x; y

=> (x + 2y + 3)² + (y + 2)² + 19 ≥ 19 với mọi x; y

=> P ≥ 19 với mọi x; y

Dấu "=" xảy ra khi x + 2y + 3 = 0 và y + 2 = 0

Bn tự giải tiếp nha, mk ko biết có nhầm chỗ nào ko nhưng cách lm như vậy đó

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\(x^2+5y^2+9z^2-4xy-6yz+12\)

\(=\left(x^2-4xy+4y^2\right)+\left(y^2-6yz+9z^2\right)+12\)

\(=\left(x-2y\right)^2+\left(y-3z\right)^2+12\ge12\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2y=0\\y-3z=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2y\\y=3z\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=6z\\y=3z\end{cases}}\)

Lời giải:

$A=(x^2+4y^2+4xy)+y^2+6x+16y+32$

$=(x+2y)^2+6(x+2y)+(y^2+4y)+32$

$=(x+2y)^2+6(x+2y)+9+(y^2+4y+4)+19$

$=(x+2y+3)^2+(y+2)^2+19\geq 0+0+19=19$

Vậy $A_{\min}=19$. Giá trị này đạt tại $x+2y+3=y+2=0$

$\Leftrightarrow y=-2; x=1$

a,<=>   x²-4x+2²+y²-8y+4²-14

<=> (x²-2x2+2²)+(y²-2x4+4²)-14

<=> (x-2)²+(y-4)²-14 

Vì (x-2)²+(y-4)²>= 0

=> F >= -14 => MIn F = -14 <=> x=2, y=4

b, <=> (x²+5²+(2y)²-4xy+10x-20y) +(y²-2y+1)+2

<=> (x+5-2y )²+(y-1)²+2 

Vì (x+5-2y) ²+(y-1)² >= 0

=> G >= 2 => Min =2 <=> y=1, x= -3

\(F=x^2-4x+y^2-8y+6\)

\(F=\left(x^2-2.2x+2^2\right)+\left(y^2-2.4.y+4^2\right)-14\)

\(F=\left(x-2\right)^2+\left(y-4\right)^2-14\)

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

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

\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\forall x\)

\(F=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy \(F_{min}=-14\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)

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

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

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

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  \(E_{Max}=-2\Leftrightarrow x=3\)