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\(A=\left(-x^2-2xy-y^2\right)-2y^2+\left(10x+10y\right)+4y-18\)
\(=-\left(x+y\right)^2+2\left(x+y\right).5-\left(2y^2-4y+2\right)-16\)
\(=-\left[\left(x+y\right)^2-2\left(x+y\right).5+5^2\right]-2\left(y-1\right)^2+9\)
\(=-\left(x+y-5\right)^2-2\left(y-1\right)^2+9\le9\forall x;y\)
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
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 \(\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\)
a, \(A=4-2x^2\le4\)
Dấu ''='' xảy ra khi x = 0
Vậy GTLN A là 4 khi x = 0
b, \(B=-x^2+10x-5=-\left(x^2-10x+5\right)=-\left(x^2-10x+25-20\right)\)
\(=-\left(x-5\right)^2+20\le20\)Dấu ''='' xảy ra khi x = 5
Vậy GTLN B là 20 khi x = 5
c, \(C=-3x^2+3x-5=-3\left(x^2-x+\frac{5}{3}\right)\)
\(=-3\left(x^2-x+\frac{1}{4}+\frac{17}{12}\right)=-3\left(x-\frac{1}{2}\right)^2-\frac{51}{12}\le-\frac{51}{21}=-\frac{17}{7}\)
Vậy GTLN C là -17/7 khi x = 1/2
d, tương tự
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 \(\Leftrightarrow\)\(\hept{\begin{cases}x+y-5=0\\y-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=4\\y=1\end{cases}}}\)