Đặt \(A=x^2+2y^2+2xy+2x+4y-1\)

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

\(A=\left[\left(x+y\right)^2+2\left(x+y\right)+1\right]+\left(y^2+2y+1\right)-3\)

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

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

Vậy GTNN của \(A\) là \(-3\) khi \(x=0\) và \(y=-1\)

Chúc bạn học tốt ~ 

Đặt \(B=-x^2-2x-y^2-8y-10\)

\(-B=\left(x^2+2x+1\right)+\left(y^2+8y+16\right)-7\)

\(-B=\left(x+1\right)^2+\left(y+4\right)^2-17\ge-17\)

\(B=-\left(x+1\right)^2-\left(y+4\right)^2+17\le17\)

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

Vậy GTLN của \(B\) là \(17\) khi \(x=-1\) và \(y=-4\)

Chúc bạn học tốt ~ 

\(Câu\text{ }1:\\ A=-2x^2-y^2-2xy+4x+2y+5\\ =-x^2-x^2-y^2-2xy+2x+2x+2y-1-1+7\\ =-\left(x^2+2xy+y^2\right)+\left(2x+2y\right)-1-\left(x^2-2x+1\right)+7\\ =-\left(x+y\right)^2+2\left(x+y\right)-1-\left(x-1\right)^2+7\\ =-\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]-\left(x-1\right)^2+7\\ =-\left(x+y-1\right)^2-\left(x-1\right)^2+7\\ =-\left[\left(x+y-1\right)^2+\left(x-1\right)^2\right]+7\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \left(x+y-1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x-1\right)^2+\left(x+y-1\right)^2\ge0\forall x;y\\ \Rightarrow-\left[\left(x-1\right)^2+\left(x+y-1\right)^2\right]\le0\forall x;y\\ \Rightarrow A=-\left[\left(x-1\right)^2+\left(x+y-1\right)^2\right]+7\le7\forall x;y\\ Dấu\text{ }"="\text{ }xảy\text{ }khi:\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(x+y-1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y+1-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\\ Vậy\text{ }A_{\left(Max\right)}=7\text{ }khi\text{ }\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)

\(Câu\text{ }2:\\ B=2x^2+4y^2+4xy+2x+4y+9\\ =x^2+x^2+4y^2+4xy+2x+4y+1+8\\ =\left(x^2+4xy+4y^2\right)+\left(2x+4y\right)+x^2+1+8\\ =\left(x+2y\right)^2+2\left(x+2y\right)+1+x^2+8\\=\left[\left(x+2y\right)^2+2\left(x+2y\right)+1\right]+x^2+8\\ =\left(x+2y+1\right)^2+x^2+8\\ Do\text{ }x^2\ge0\forall x\\ \left(x+2y+1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x+2y+1\right)^2+x^2\ge0\forall x;y\\ \Rightarrow\left(x+2y+1\right)^2+x^2+8\ge8\forall x;y\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\left\{{}\begin{matrix}x^2=0\\\left(x+2y+1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.\\ Vậy\text{ }B_{\left(Min\right)}=8\text{ }khi\text{ }\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right. \)

\(\)

Chữa đề: \(A=-2x^2-y^2-2xy+4x+2y+5\)

Ở

QWERTY huhu

\(A=-x^2+2xy-4y^2+2x+10y-8\)

\(=-x^2+2xy-y^2-3y^2+2x-2y+12y-12+4\)

\(=-\left(x^2-2xy+y^2\right)+\left(2x-2y\right)-1-\left(3y^2-12y+12\right)+5\)

\(=-\left(x-y\right)^2+2\left(x-y\right)-1-3\left(y-2\right)^2+5\)

\(=-\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]\)\(-3\left(y-2\right)^2+5\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+5\)

\(A_{max}=5\Leftrightarrow\hept{\begin{cases}\left(x-y-1\right)^2=0\\3\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}}}\)

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

\(KL:A_{max}=5\Leftrightarrow x=3;y=2\)

a) A= -x² + 6x -10

       = -(x² - 6x) -10

       =  -(x² - 2. x .3 +3² -9)- 10

      = -( x-3 )²  +9 -10 

      = - (x-3)² -1 \(\le\)-1 với mọi giá trị của x

       Dấu '' = '' xảy ra khi và chỉ khi

               x-3 =0

               \(\Leftrightarrow\)x=3

Vậy giá trị lớn nhất của biểu thức A là -1 tại x =3

CÁC PHẦN KHÁC CẬU LÀM TƯƠNG TỰ

b) B= -2x²-4x-10

        = -2(x²+ 2x ) -10

        = -2 (x² +2x+1² -1)-10

         =-2(x+1)² +2 -10

        =-2(x+1)² -8  \(\le\)-8 với mọi giá trị của x

Dấu " ='' xảy ra khi và chỉ khi

        x+1=0

............................

c) C= -2x² +3x -10

       = -2(x² -\(\frac{3}{2}\)x) -10

       = -2( x² - 2.x.\(\frac{3}{4}\)+ \(\frac{3^2}{4^2}\)-\(\frac{9}{16}\))-10

       = -2(x-\(\frac{3}{4}\))² +\(\frac{9}{8}\)-10

        =-2(x- \(\frac{3}{4}\))² +\(\frac{-71}{8}\)\(\le\)\(\frac{-71}{8}\)với mọi giá trị của x

Dấu  bằng ''='' xảy ra khi và chi khi  

     x-\(\frac{3}{4}\)=0

   .......................................................

d)  D= -x² -y²+2x-4y -10

          =(-x²+2x) +( -y² -4y) -10

          = -(x² -2x+1 -1) -(y² +4y+2²-4 )-10 

          =-(x-1)² +1  -(y+2)² +4 -10

           =-(x-1)² - (y+2)² -5   \(\le\)5  với mọi giá tri của x

Dấu '' ='' xảy ra khi và chỉ khi  

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

......................................................

e) XIN LỖI TỚ CHƯA NGHĨ RA

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

a)\(M=x^2-2xy+2y^2-4y+2016\)

\(=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+2012\)

\(=\left(x-y\right)^2+\left(y-2\right)^2+2012\ge2012\)

Dấu = khi \(\begin{cases}\left(x-y\right)^2=0\\\left(y-2\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y=0\\y-2=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=y\\y=2\end{cases}\)\(\Leftrightarrow x=y=2\)

Vậy Min_M=2012 khi x=y=2

b)\(N=x^2-2xy+2x+2y^2-4y+2016\)

\(=\left(x^2-2xy+2x+y^2-2y+1\right)+\left(y^2-2y+1\right)+2014\)

\(=\left(x-y+1\right)^2+\left(y-1\right)^2+2014\ge2014\)

Dấu = khi \(\begin{cases}\left(x-y+1\right)^2=0\\\left(y-1\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y+1=0\\y-1=0\end{cases}\)

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

Vậy Min_N=2014 khi x=0;y=1