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transformers123

$x^4+y^4+(x^2+y^2-2)(2xy-1)+3x^2y^2-1=0$

$\iff (x^2+y^2)^2+(x^2+y^2-2)(2xy-1)+x^2y^2-1=0\ \bigstar$

Đặt $x^2+y^2=a,\ xy=b\ (a \ge 2b)$

Từ $\bigstar \Longrightarrow a^2+(a-2)(2b-1)+b^2-1=0$

$\iff a^2+2ab-a-4b+b^2+1=0$

Ta có: $a^2+2ab-a-4b+b^2+1 \ge a^2+2ab-2a-2b+b^2+1$ (vì $a \ge 2b$)

$\iff a^2+2ab-a-4b+b^2+1 \ge (a+b-1)^2$

$\iff a^2+2ab-a-4b+b^2+1 \ge 0$

Dấu "=" xảy ra khi:

$\begin{cases}a=2b\\a+b-1=0\end{cases} \Longrightarrow
\begin{cases}x^2+y^2=2xy\\x^2+y^2+xy-1=0\end{cases} \iff \begin{cases}(x-
y)^2=0\\x^2+y^2+xy-1=0\end{cases}$

$\iff \begin{cases}x=y\\x^2+x^2+x^2-1=0\end{cases} \iff \begin{cases}x=y\\x= \pm
\dfrac{\sqrt{3}}{3}\end{cases} \iff \begin{bmatrix}x=y=\dfrac{\sqrt{3}}{3}\\x=y=-\dfrac{\sqrt{3}}{3}\end{bmatrix}$

Vậy $(x;y)=(\dfrac{\sqrt{3}}{3};\dfrac{\sqrt{3}}{3}),\ (-\dfrac{\sqrt{3}}{3};-\dfrac{\sqrt{3}}{3})$
 
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