\(\left\{{}\begin{matrix}x^3+y^3=1\\x^2+y^5=x^5+y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2-xy\right)=1\left(1\right)\\x^2-y^2-\left(x^5-y^5\right)=0\left(2\right)\end{matrix}\right.\)
Đặt s = x + y ; p = xy \(\left(s^2\ge4p\right)\)
(1) \(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=1\) hay \(s.\left(s^2-3p\right)=1\)
(2) \(\Leftrightarrow\left(x-y\right)\left[x+y-\left(x^4+x^3y+yx^3+y^4+x^2y^2\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\\left(x+y\right)-\left[\left(x^2+y^2\right)^2-x^2y^2+xy\left(x^2+y^2\right)\right]=0\left(3\right)\end{matrix}\right.\)
Với x = y ; ta có : \(2x^3=1\Leftrightarrow x^3=\dfrac{1}{2}\) \(\Leftrightarrow x=\sqrt[3]{\dfrac{1}{2}}=y\) (t/m)
(3) \(\Leftrightarrow s-\left[\left(s^2-2p\right)^2-p^2+p\left(s^2-2p\right)\right]=0\)
\(\Leftrightarrow s-\left[s^4-3ps^2+p^2\right]=0\) \(\Leftrightarrow s-\left[s^2\left(s^2-3p\right)+p^2\right]=0\)
Nếu s = 0 thì x = -y thay vào (1) ko t/m . Do đó : s \(\ne0\)
Ta có : \(s^2-3p=\dfrac{1}{s}\) . Khi đó : \(-p^2=0\Leftrightarrow p=0\) hay xy = 0 \(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
x = 0 thay vào (1) thì y = 1
y = 0 thay vào (1) thì x = 1
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