Tìm GTLN của:
\(A=-x^2+2xy-4y^2+2x+10y-3\)
Tìm GTLN C= -x^2 + 2xy - 4y^2 + 2x +10y -3
\(C=-x^2+2xy-4y^2+2x+10y-3\)
\(=-\left(x^2+2xy-y^2\right)+2x-2y-1-3y^2+12y-12+10\)
\(=-\left(x-y\right)^2+2\left(x-y\right)-1-3\left(y^2-4y+4\right)+10\)
\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+10\le10\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}}\)
Vậy \(C_{max}=10\) tại x = 3; y = 2
tìm gtln của -x^2+2xy-4y^2+2x+10y-8
Tìm GTLN của:
\(A=-x^2+2xy-4y^2+2x+10y-8\)
\(A=-x^2+2xy-4y^2+2x+10y-8\)
\(=-x^2+2xy-y^2-3y^2+2x-2y+12y-12+4\)
\(=-\left(x^2-2xy+y^2\right)+\left(2x-2y\right)-1-\left(3y^2-12y+12\right)+5\)
\(=-\left(x-y\right)^2+2\left(x-y\right)-1-3\left(y-2\right)^2+5\)
\(=-\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]\)\(-3\left(y-2\right)^2+5\)
\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+5\)
\(A_{max}=5\Leftrightarrow\hept{\begin{cases}\left(x-y-1\right)^2=0\\3\left(y-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}x-y-1=0\\y=2\end{cases}}\)\(\Rightarrow x-2-1=0\Leftrightarrow x=3\)
\(KL:A_{max}=5\Leftrightarrow x=3;y=2\)
tìm GTLN của biểu thức
-x^2+2xy-4y^2+2x+10y-3
nhanh mk cần gấp
tìm GTLN: -x^2+2xy-4y^2+2x+10y-8
\(A=-x^2+2xy-4y^2+2x+10y-8\)
\(=-\left(x^2-2xy+4y^2-2x-10y+8\right)\)
\(=-\left[\left(x-y-1\right)^2+3\left(y-2\right)^2-5\right]\)
\(=5-\left(x-y-1\right)^2-3\left(y-2\right)^2\le5\)
Dấu"=" xảy ra <=> \(\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}}\) <=> \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)
Vậy MAX \(A=5\)khi \(x=3;\)\(y=2\)
Tìm GTLN của:
a.M= 2+x-x2
b.S= -x2+2xy-4y2+2x+10y-3
a ) \(M=2+x-x^2\)
\(=-x^2+x-\frac{1}{4}+\frac{9}{4}\)
\(=-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\)đạt GTNN là \(\frac{9}{4}\) tại x = \(\frac{1}{2}\)
b ) \(S=-x^2+2xy-4y^2+2x+10y-3\)
\(=\left[\left(-x^2+2xy-y^2\right)+\left(2x-2y\right)-1\right]+\left(-3y^2+12y-12\right)+10\)
\(=\left[-\left(x-y\right)^2+2\left(x-y\right)-1\right]-3\left(y-2\right)^2+10\)
\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+10\le10\) có GTLN là 10
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\y-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=2\end{cases}}}\)
Vậy \(S_{max}=10\Leftrightarrow x=3;y=2\)
Tìm \(x,\) \(y\) sao cho:
\(B=-x^2+2xy-4y^2+2x+10y-8\) có \(GTLN\)
Tìm GTLN của biểu thức :
a) A = 5 - 8x - 2x^2
b) B = -x^2 + 2xy - 4y^2 + 2x + 10y - 9
a)\(A=5-8x-2x^2\)
\(=-2\left(x^2+4x-\frac{5}{2}\right)\)
\(=-2\left(x^2+4x+4-\frac{13}{2}\right)\)
\(=-2\left[\left(x+2\right)^2-\frac{13}{2}\right]\)
\(=-2\left[\left(x+2\right)^2\right]+13\le13\)
Vậy \(A_{max}=13\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Tìm GTLN của
A= -x2 +2xy - 4y2 + 2x + 10y +5
B= -x2 - 2y2 -2xy + 2x - 2y -15