Ta có: \(A=2\left(x^3+y^3\right)-3xy=2\left(x+y\right)\left(x^2-xy+y^2\right)-3xy\)
Lại có: \(x^2+y^2=2\Leftrightarrow\left(x+y\right)^2-2xy=2\Leftrightarrow xy=\dfrac{\left(x+y\right)^2-2}{2}\)
\(\Rightarrow A=2\left(x+y\right)\left(2-\dfrac{\left(x+y\right)^2-2}{2}\right)-\dfrac{3\left(x+y\right)^2-2}{2}\)
Đặt \(t=x+y\Rightarrow\left|t\right|\le2\) và \(A=-t^3-\dfrac{3}{2}t^2+6t+3\forall\left|t\right|\le2\)
\(\Rightarrow g'\left(t\right)=-3t^2-3t+6\)
\(g'\left(t\right)=0\Rightarrow-3t^2-3t+6=0\)
\(\Rightarrow\left(t-1\right)\left(t+2\right)=0\)\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-2\end{matrix}\right.\)\(t\in\left[-2;2\right]\)
\(g\left(-2\right)=-7;g\left(2\right)=1;g\left(1\right)=\dfrac{13}{2}\)
Nhìn vào các số trên rõ ràng là \(A_{MAX}=\dfrac{13}{2}\Leftrightarrow x=\dfrac{1\pm\sqrt{3}}{2};y=\dfrac{1\mp\sqrt{3}}{2}\)
\(A_{Min}=-7\Leftrightarrow x=y=-1\)
GTLN:
áp dụng BĐT Cauchy-Swarch: \(\left(x^2+y^2\right)\left(1^2+1^2\right)\ge\left(x+y\right)^2\)
\(\Rightarrow x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)\(\Rightarrow\left(x+y\right)^2\le4\Rightarrow-2\le x+y\le2\)
ta có: \(A=2\left(x+y\right)\left(x^2-xy+y^2\right)-3xy=2\left(x+y\right)\left(2-xy\right)-3xy\)
mà \(x+y\le2\Rightarrow A\le4\left(2-xy\right)-3xy=8-7xy\)
mà \(x^2+y^2=2\Rightarrow\left(x+y\right)^2-2=2xy\Rightarrow\dfrac{7}{2}\left(x+y\right)^2-7=7xy\)
\(\Rightarrow-\dfrac{7}{2}\left(x+y\right)^2+7+8=8-7xy\)
\(\Rightarrow-\dfrac{7}{2}\left(x+y\right)^2+15=8-7xy\)
\(\Rightarrow A\le15-\dfrac{7}{2}\left(x+y\right)^2\le15\)
\(\Rightarrow MaxA=15\) khi \(\left[{}\begin{matrix}x=1;y=-1\\x=-1;y=1\end{matrix}\right.\)
GTNN:
áp dụng BĐT \(a^3+b^3+c^3\ge3abc\)
\(\Rightarrow x^3+y^3+1\ge3xy\Rightarrow x^3+y^3\ge3xy-1\)
\(\Rightarrow A=2\left(x^3+y^3\right)-3xy\ge2\left(3xy-1\right)-3xy=3xy-2\)
do \(x^2+y^2=2\Rightarrow\left(x+y\right)^2-2=2xy\Rightarrow\dfrac{3}{2}\left(x+y\right)^2-3=3xy\)
\(\Rightarrow A\ge\dfrac{3}{2}\left(x+y\right)^2-3-2=\dfrac{3}{2}\left(x+y\right)^2-5\ge-5\)
\(\Rightarrow A\ge-5\) Vậy \(MinA=-5\)