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

dấu = xảy ra \(\Leftrightarrow x=-\frac{2}{5}\)

Bài 1:

a)\(F=x^2+26y^2-10xy+14x-76y+59\)

         \(=\left(x^2-2\cdot x\cdot5y+25y^2\right)+\left(14x-70y\right)+\left(y^2-6x+9\right)+50\)

        \(=[\left(x-5y\right)^2+14\left(x-5y\right)+49]+\left(y-3\right)^2+1\)

          \(=\left(x-5y+7\right)^2+\left(y-3\right)^2+1\ge1\)

 Để Fmin=1 thì y=3;x=8

b)\(H=m^2-4mp+5p^2+10m-22p+28\)

         \(=\left(m^2-2\cdot m\cdot2p+4p^2\right)+\left(10m-20p\right)+\left(p^2-2p+1\right)+27\)

         \(=[\left(m-2p\right)^2+2\cdot\left(m-2p\right)\cdot5+25]+\left(p-1\right)^2+2\)

           \(=\left(m-2p+5\right)^2+\left(p-1\right)^2+2\ge2\)

Để Hmin=2 thì p=1;m=-3

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

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

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

Bmax=1 khi x+y=0 và y-1=0=>x=-1;y=1

2/C=\(x^2+x+\frac{1}{4}+y^2+y+\frac{1}{4}+\frac{1}{2}\)

      =\(\left(x+\frac{1}{2}\right)^2+\left(y+\frac{1}{2}\right)^2+\frac{1}{2}\)>=\(\frac{1}{2}\)

Cmin=\(\frac{1}{2}\)khi \(x+\frac{1}{2}=0\)và \(y+\frac{1}{2}=0\)=>\(x=y=\frac{-1}{2}\)

\(B=-x^2-2y^2-2xy+2y\)

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

\(=-\left(x+y\right)^2-\left(y-1\right)^2+1\le1\) đạt GTLN là 1

Khi x = - 1; y = 1

cảm ơn bạn Đinh Đức Hùng nha

a) \(A=4x^2-12x+100=\left(2x\right)^2-12x+3^2+91=\left(2x-3\right)^2+91\)

Ta có: \(\left(2x-3\right)^2\ge0\forall x\inℤ\)

\(\Rightarrow\left(2x-3\right)^2+91\ge91\)

hay A \(\ge91\)

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

<=> 2x-3=0

<=> 2x=3

<=> \(x=\frac{3}{2}\)

Vậy Min A=91 đạt được khi \(x=\frac{3}{2}\)

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

Ta có: \(-\left(x+\frac{1}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\) hay B\(\le\frac{5}{4}\)

Dấu "=" \(\Leftrightarrow-\left(x+\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x+\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy Max B=\(\frac{5}{4}\)đạt được khi \(x=\frac{-1}{2}\)

\(C=2x^2+2xy+y^2-2x+2y+2\)

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

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

Ta có: 

\(\hept{\begin{cases}\left(x+y-1\right)^2\ge0\forall x;y\inℤ\\x^2\ge0\forall x\inℤ\end{cases}}\)

\(\Leftrightarrow\left(x+y-1\right)^2+x^2+1\ge1\)

hay C\(\ge\)1

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

Vậy Min C=1 đạt được khi y=1 và x=0