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\(A=-x^2-5y^2+2xy-4x+20y+13\)
\(=-x^2+2xy-y^2-4y^2-4x+4y+16y+13\)
\(=-\left(x^2-2xy+y^2\right)-\left(4y^2-16y+16\right)-\left(4x-4y\right)+29\)
\(=-\left(x-y\right)^2-4\left(y-2\right)^2-4\left(x-y\right)-4+25\)
\(=-\left[\left(x-y\right)^2+4\left(x-y\right)+4\right]-4\left(y-2\right)^2+25\)
\(=-\left(x-y+2\right)^2-4\left(y-2\right)^2+25\)
\(A_{max}=25\Leftrightarrow\hept{\begin{cases}\left(x-y+2\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y+2=0\\y=2\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=0\\y=2\end{cases}}\)
\(B=-7x^2-y^2+4xy+16x-2y+17.\)
\(=-4x^2+4xy-y^2-3x^2+12x-12+4x-2y+29\)
\(=-\left(2x-y\right)^2-3\left(x-2\right)^2+2\left(2x-y\right)^2-1+30\)
\(=-\left[\left(2x-y\right)^2-2\left(2x-y\right)^2+1\right]-3\left(x-2\right)^2+30\)
\(=-\left(2x-y-1\right)^2-3\left(x-2\right)^2+30\)
\(\Rightarrow B_{max}=30\Leftrightarrow\hept{\begin{cases}\left(2x-y-1\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x-y-1=0\\x=2\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)
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
Ta có: 5x2+10y2-6xy-4x-2y +3= x2 -6xy +(3y)2 +4x2 +y2 -4x -2y +3
= (x - 3y)2 +(2x)2 -4x+1+ y2 -2y+1 +1
= (x-3y)2 + (2x -1)2 + (y-1)2 +1
Ta có :(x-3y)2 luôn lớn hơn hoặc bằng 0
(2x -1)2 luôn lớn hơn hoặc bằng 0
(y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 luôn lớn hơn hoặc bằng 0
=>(x-3y)2 + (2x -1)2 + (y-1)2 +1 >0
ta có:\(A=x^2+5y^2-4xy-2y+2x+2010\)
\(=x^2+4y^2+y^2-4xy-4y+2y+2x+1+1+2008\)
\(=\left(x^2-4xy+4y^2\right)+\left(2x-4y\right)+1+\left(y^2+2x+1\right)+2008\)
\(=\left(x-2y\right)^2+2\left(x-2y\right)+1+\left(y+1\right)^2+2008\)
\(=\left(x-2y+1\right)^2+\left(y+1\right)^2+2008\)
Vì: (x-2y+1)2+(y+1)>0 với \(\forall x;y\)
do đó: (x-2y+1)2+(y+1)+2008 > 2008 với \(\forall x;y\)
Dấu "=" xảy ra khi x-2y+1=0 và y+1=0
ta có:
y+1=0=>y=0-1=>y=-1
thay y=-1 và x-2y+1=0
=>x-2.(-1)+1=0
=>x+2+1=0
=>x+2=-1
=>x=-1-2
=>x=-3
vậy \(A_{min}=2008\) khi x=-3 hoặc x=-1
Có x^2 + 2xy + 4x + 4y + 2y^2 + 3 = 0
--> (x+y)^2 + 4(x+y) + 4+ y^2 - 1 = 0
--> (x+y+2)^2 + y^2 = 1
-->(x+y+2)^2 <= 1 ( vì y^2 >=1)
--> -1 <= x+y+2 <=1
--> 2015 <= x+y+2018 <= 2017
hay 2015 <= Q , dau bang xay ra khi x+y+2=-1 --> x+y=-3
Q<=2017, dau bang xay ra khi x+y+2=1 --> x+y=-1
Vậy giá trị nhỏ nhất của Q là 2015 khi x+y =-3
giá trị lớn nhất của Q là 2017 khi x+y=-1
\(A=x^2+4y^2-2xy+4x-10y+2020.\)
\(=\left(x^2-2xy+y^2\right)+\left(3y^2-6y+3\right)+\left(4x-4y\right)+2017\)
\(=\left(x-y\right)^2+3\left(y-1\right)^2+4\left(x-y\right)+2017\)
\(=\left[\left(x-y\right)^2+4\left(x-y\right)+4\right]+3\left(y-1\right)^2+2013\)
\(=\left(x-y+2\right)^2+3\left(y-1\right)^2+2013\)
\(A_{min}=2013\Leftrightarrow\hept{\begin{cases}\left(x-y+2\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-y+2=0\\y=1\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}}\)
\(B=8x^2+y^2-4xy-12x+2y+30\)
\(=\left(4x^2-4xy+y^2\right)+\left(4x^2-8x+4\right)-\left(4x-2y\right)+26\)
\(=\left(2x-y\right)^2+4\left(x-1\right)^2-2\left(2x-y\right)+26\)
\(=\left[\left(2x-y\right)^2-2\left(2x-y\right)+1\right]+4\left(x-1\right)^2+25\)
\(=\left(2x-y-1\right)^2+4\left(x-1\right)^2+25\)
\(\Rightarrow B_{min}=25\)\(\Leftrightarrow\hept{\begin{cases}\left(2x-y-1\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x-y-1=0\\x=1\end{cases}}\)\(\Leftrightarrow x=y=1\)
\(A=-\left(4x^2-4xy+y^2\right)-\left(y^2-2y+1\right)+4\)
\(A=4-\left(2x-y\right)^2-\left(y-1\right)^2\le4\)
\(A_{max}=4\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=1\end{matrix}\right.\)