cho x^2+y^2-xy=1
P=x^4+y^4-x^2.y^2 tim min,maxP:xcac ban giup minh giai bai nay nha minh dang can gap.thanks
[tex]Am-Gm \Rightarrow 1=x^2+y^2-xy \geq 2xy-xy=xy[/tex]
[tex]1=x^2+y^2-yz=(x+y)^2-3xy \Rightarrow xy \geq -\frac{1}{3}[/tex]
[tex]\Rightarrow 1 \geq xy \geq \frac{-1}{3}[/tex]
[tex](gt) \Rightarrow x^2+y^2=xy+1[/tex]
[tex]\Rightarrow x^4+y^4+x^2+y^2=2xy+1[/tex]
[tex]\Rightarrow x^4+y^4-x^2y^2=-2x^2y^2+2xy+1[/tex]
[tex]Dat:t=xy \Rightarrow x^4+y^4-x^2y^=-2t^2+2t+1[/tex]
[tex]Xet: f(t)=-2t^2+2t+1; \forall t \in\ [-\frac{1}{3};1][/tex]
[tex]BBT \Rightarrow max f(t)=f(\frac{1}{2})=\frac{3}{2}[/tex]
Đẳng thức xảy ra [tex]\Leftrightarrow \left{\begin{x^2+y^2-xy=1}\\{xy=\frac{1}{2}}[/tex]
[tex]\Rightarrow \left{\begin{xy=\frac{1}{2}}\\{x^2+y^2=\frac{3}{2}[/tex]
[tex]Viete \Rightarrow \left[\begin{x=\frac{\sqrt{5} \pm \ 1}{2.\sqrt{2}},y=\frac{\sqrt{5} \pm \ 1}{2.\sqrt{2}}\\{x=-\frac{\sqrt{5}\pm \ 1}{2.\sqrt{2}};y=-\sqrt{5} \pm \ 1}{2.\sqrt{2}}}[/tex]
[tex]min f(t)=f(-\frac{1}{3})=\frac{1}{9} [/tex]
Đẳng thức xảy ra [tex]\Leftrightarrow \left{\begin{x^2+y^2=\frac{2}{3}}\\{xy=-\frac{1}{3}}[/tex]
[tex]Viete \Rightarrow \left{\begin{x=\pm\ \frac{\sqrt{3}}{3}}\\{y=\pm \ \frac{\sqrt{3}}{3}}[/tex]
P/s: rãnh lục lại post chơi.
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