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A = -x2 - 3y2 - 2xy + 10x + 14y - 18
A = -x2 - y2 -25 + 10x +10y -2xy -2y2 + 4y -2 + 9
A = -(x2 + y2 + ( -5 )2 - 10x - 10y + 2xy ) - 2 (y2 - 2y + 1 ) + 9
A = -( x + y - 5 )2 - 2 ( y - 1 )2 + 9
-( x + y - 5 )2 \(\le\)0 ; - 2 ( y - 1 )2 \(\le\)0
\(\Rightarrow\)A \(\le\)0 + 0 + 9 = 9
Dấu " = " xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+y-5=0\\y-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\y=1\end{cases}}}\)
\(A=\left(-x^2-2xy-y^2\right)-2y^2+\left(10x+10y\right)+4y-18\)
\(=-\left(x+y\right)^2+2\left(x+y\right).5-\left(2y^2-4y+2\right)-16\)
\(=-\left[\left(x+y\right)^2-2\left(x+y\right).5+5^2\right]-2\left(y-1\right)^2+9\)
\(=-\left(x+y-5\right)^2-2\left(y-1\right)^2+9\le9\forall x;y\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y-5=0\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5-y\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\y=1\end{cases}}\)
Vậy \(A_{max}=9\Leftrightarrow\hept{\begin{cases}x=4\\y=1\end{cases}}\)
-2A=2x2+6y2+4xy-20x-28y+36
=(x2+4xy+4y2)+(x2-20x+100)+2(y2-14y+49)-162
=(x+2y)2+(x-10)2+2(y-7)2-162\(\ge\)-162
=> A\(\le81\)
Dấu "=" xảy ra khi
Làm nốt phần còn lại của bạn Thắng
(x + y - 5)2 + 2(y - 1)2 - 9 = 0
<=> 2(y - 1)2 = 9 - (S - 5)2 \(\ge0\)
\(\Leftrightarrow\left(S-5\right)^2\le9\)
\(\Leftrightarrow-3\le S-5\le3\)
\(\Leftrightarrow2\le S\le8\)
Vậy GTNN là 2 đạt được khi x = y = 1
GTLN là 8 đạt được khi (x, y) = (7, 1)
\(x^2+3y^2+2xy-10x-14y+18\)
\(\Rightarrow\left(x^2+2xy-10x+y^2-10y+25\right)+2y^2-4y-7=0\)
\(\Rightarrow\left(x+y-5\right)^2+2y^2-4y+2-9=0\)
\(\Rightarrow\left(x+y-5\right)^2+2\left(y^2-2y+1\right)-9=0\)
\(\Rightarrow\left(x+y-5\right)^2+2\left(y-1\right)^2-9=0\)
....
gợi ý nhé:
[-(x-y)2-10(x-y)-25] - 2(y-1)2 + 2010
= -[(x-y)+5]2 - 2(y-1)2 + 2010
tự cậu suy ra MAX nhé
chưa hiểu thì hỏi nhé
\(A=-x^2-3y^2-2xy+10x+14y-18\\ =-x^2-y^2-2y^2-2xy+10x+10y+4y-25-2+9\\ =-\left(x^2+y^2+25+2xy-10x-10y\right)-\left(2y^2-4y+2\right)+9\\ \\ =-\left(x+y-5\right)^2-2\left(y^2-2y+1\right)+9\\ =-\left(x+y-5\right)^2-2\left(y-1\right)^2+9\)Do \(-\left(x+y-5\right)^2\le0\forall x;y\)
\(-2\left(y-1\right)^2\le0\forall y\)
\(\Rightarrow-\left(x+y-5\right)^2-2\left(y-1\right)^2\le0\forall x;y\)
\(\Rightarrow A=-\left(x+y-5\right)^2-2\left(y-1\right)^2+9\le9\forall x\)
Dấu "='' xảy ra khi: \(\left\{{}\begin{matrix}-\left(x+y-5\right)^2=0\\-2\left(y-1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5-y\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\)
Vậy \(A_{\left(Max\right)}=9\) khi \(\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\)
Ta có :
\(x^2+3y^2+2xy-10x-14y+18=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)-10x-10y+25+\left(2y^2-4y+2\right)-9=0\)
\(\Leftrightarrow\left(x+y\right)^2-2.\left(x+y\right).5+25+2\left(y^2-2y+1\right)=9\)
\(\Leftrightarrow\left(x+y-5\right)^2+2\left(y-1\right)^2=9\)
Vì \(2\left(y-1\right)^2\ge0\forall y\)nên \(\left(x+y-5\right)^2\le9\)hay \(\left(M-5\right)^2\le9\)
\(\Rightarrow-3\le M-5\le3\Leftrightarrow2\le M\le8\)
- \(Min_M=2\)khi \(\hept{\begin{cases}x=1\\y=1\end{cases}}\)
- \(Max_M=8\)khi\(\hept{\begin{cases}x=7\\y=1\end{cases}}\)
Lời giải:
\(x^2+3y^2+10x-14y-2xy=11\)
$\Leftrightarrow (x^2-2xy+y^2)+2y^2+10x-14y=11$
$\Leftrightarrow (x-y)^2+10(x-y)+25+(2y^2-4y+2)=38$
$\Leftrightarrow (x-y+5)^2+2(y-1)^2=38$
$\Rightarrow (x-y+5)^2=38-2(y-1)^2\leq 38$
$\Rightarrow -\sqrt{38}\leq x-y+5\leq \sqrt{38}$
$\Leftrightarrow -\sqrt{38}-5\leq x-y\leq \sqrt{38}-5$
Vậy $A_{\min}=-\sqrt{38}-5$ và $A_{\max}=\sqrt{38}-5$
\(A=-\left(x^2+y^2+25+2xy-10x-10y\right)-2y^2+4y-2+9\)
\(A=-\left(x+y-5\right)^2-2\left(y-1\right)^2+9\le9\)
\(\Rightarrow A_{max}=9\) khi \(\left\{{}\begin{matrix}y=1\\x=4\end{matrix}\right.\)
\(A_{min}\) không tồn tại