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\(P-\dfrac{5}{2}=x+2y-\dfrac{x^2+y^2}{2}=-\dfrac{1}{2}\left(x-1\right)^2-\dfrac{1}{2}\left(y-2\right)^2+\dfrac{5}{2}\le\dfrac{5}{2}\)
\(\Rightarrow P-\dfrac{5}{2}\le\dfrac{5}{2}\Rightarrow P\le5\)
\(P_{max}=5\) khi \(\left(x;y\right)=\left(1;2\right)\)
P = x6 + y6 = (x2 + y2)(x4 - x2 y2 + y4)
= (x2 + y2)2 - 3x2 y2 \(\ge1-3×\frac{\left(x^2+y^2\right)^2}{4}=1-\frac{3}{4}=\frac{1}{4}\)
Đạt được khi x2 = y2 = \(\frac{1}{2}\)
\(P=\dfrac{x+2y}{2xy}+\dfrac{1}{x+2y}=\dfrac{x+2y}{4}+\dfrac{1}{x+2y}\)
\(P=\dfrac{x+2y}{16}+\dfrac{1}{x+2y}+\dfrac{3\left(x+2y\right)}{16}\)
\(P\ge2\sqrt{\dfrac{x+2y}{16\left(x+2y\right)}}+\dfrac{3}{16}.2\sqrt{2xy}=\dfrac{5}{4}\)
\(P_{min}=\dfrac{5}{4}\) khi \(\left(x;y\right)=\left(2;1\right)\)
Ta có : (x+y)2+7x+7y+y2+6=0
( x2 + y2 + \(\frac{49}{4}\)+ 7x + 7y + 2xy ) + y2 - \(\frac{25}{4}\)= 0
( x + y + \(\frac{7}{2}\))2 = \(\frac{25}{4}\)- y2 \(\le\frac{25}{4}\)
\(\Rightarrow\frac{-5}{4}\le x+y+\frac{7}{2}\le\frac{5}{4}\)
\(\Rightarrow\frac{-15}{4}\le x+y+1\le\frac{-5}{4}\)
\(\Rightarrow\)......
lon so roi,
thay -5/4 thành -5/2 ; 5/4 thành 5/2
-15/4 thành -5 ; 5/2 thành 0
Ta có \(x^2+y^2+xy+x=y-1\)
\(\Leftrightarrow2x^2+2y^2+2xy+2x-2y+2=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
\(\Rightarrow B=\left(-1+1-1\right)^{2023}\) \(=\left(-1\right)^{2023}\) \(=-1\)
