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

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

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

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

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

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

Đề thiếu rồi bạn ơi

=0 nữa nha cảm ơn bạn nhanh nhanh giúp mình chút:3

a)

Ta có:

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

\(\ge0-2=-2\)

Vậy \(A_{min}=-2\), đạt được khi và chỉ khi \(x-1=0\Leftrightarrow x=1\)

b)\(B=4x^2+4x+8=4x^2+4x+1+7\)

\(=\left(2x+1\right)^2+7\ge0+7=7\)

Vậy \(B_{min}=7\), đạt được khi và chỉ khi \(2x+1=0\Leftrightarrow x=\dfrac{-1}{2}\)

c)

Ta có:

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

\(=2+\dfrac{9}{4}-\left(x^2-2x.\dfrac{3}{2}+\dfrac{9}{4}\right)\)

\(=\dfrac{17}{4}-\left(x-\dfrac{3}{2}\right)^2\le\dfrac{17}{4}-0=\dfrac{17}{4}\)

Vậy \(C_{max}=\dfrac{17}{4}\), đạt được khi và chỉ khi \(x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

d) Ta có:

\(D=-x^2-5x=-\left(x^2+5x\right)=\dfrac{25}{4}-\left(x^2+2x.\dfrac{5}{2}+\dfrac{25}{4}\right)\)

\(=\dfrac{25}{4}-\left(x+\dfrac{5}{2}\right)^2\le\dfrac{25}{4}-0=\dfrac{25}{4}\)

Vậy \(D_{max}=\dfrac{25}{4}\), đạt được khi và chỉ khi \(x+\dfrac{5}{2}=0\Leftrightarrow x=-\dfrac{5}{2}\)

e) Ta có:

\(E=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2+4y^2+5^2-4xy+10x-20y+y^2-2y+1+2\)

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

\(\ge0+0+2=2\)

Vậy \(E_{min}=2\), đạt được khi và chỉ khi \(x-2y+5=y-1=0\)

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

biến đổi tương đương A = \((x^2-6x+9)+(y^2-22y+121)+(z^2+12z+36)\)\(+2019\)

=> A = \((x-3)^2+(y-11)^2+(z+6)^2+2019\ge2019\)

VẬY GTNN CỦA A LÀ 2019 ĐẠT ĐƯỢC TẠI x=3 , y=11,z=-6

H=\(x^6-2x^3+x^2-2x+2\)

\(=x^6+2x^5+3x^4+2x^2-2x^5-4x^4-6x^3-4x^2-4x+x^4+2x^3+3x^2+2x+2\)

\(=x^2\left(x^4+2x^3+3x^2+2\right)-2x\left(x^4+2x^3+3x^2+2\right)+\left(x^4+2x^3+3x^2+2\right)\)

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

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

\(=\left(x-1\right)^2\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]\text{≥}0\)

Vì \(\left\{{}\begin{matrix}\left(x-1\right)^2\text{≥}0\\\left(x^2+1\right)\text{≥}1\\\left(x+1\right)^2+1\text{≥}1\end{matrix}\right.\)

⇒ MinH=0 ⇔ \(x=1\)

\(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=3\\ B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\\ B_{min}=51\Leftrightarrow x=5\\ C=\left[\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25\right]+\left(y^2-2y+1\right)+2\\ C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\\ C_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-5=2-5=-3\\y=1\end{matrix}\right.\)

a) \(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)

\(minA=2\Leftrightarrow x=3\)

b) \(B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\)

\(minB=51\Leftrightarrow x=5\)

c) \(C=\left[x^2-2x\left(2y-5\right)+\left(2y-5\right)^2\right]+\left(y^2-2y+1\right)+2=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

\(minC=2\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

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

\(2x^2+9y^2-6xy-6x-12y+2004\)

\(=x^2-10x+25+x^2+9y^2+4-6xy+4x-12y+1975\)

\(=\left(x-5\right)^2+\left(x-3y+2\right)^2+1975\ge1975\)

Dấu \(=\)khi \(\hept{\begin{cases}x-5=0\\x-3y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}\).