Question

Giải hpt :\(\left\{{}\begin{matrix}x^3+1=2y\\y^3+1=2x\end{matrix}\right.\)

Help me !!!!!

Answer 1

Lời giải:

Lấy PT(1) trừ PT(2) ta có:

\(x^3-y^3=2y-2x=-2(x-y)\)

\(\Leftrightarrow (x-y)(x^2+xy+y^2)+2(x-y)=0\)

\(\Leftrightarrow (x-y)(x^2+xy+y^2+2)=0\)

\(\Rightarrow \left[\begin{matrix} x-y=0\\ x^2+xy+y^2+2=0\end{matrix}\right.\)

Nếu $x-y=0\Rightarrow x=y$

Thay vào PT(1): \(x^3+1=2y=2x\)

\(\Leftrightarrow x^3-2x+1=0\Leftrightarrow (x-1)(x^2+x-1)=0\)

\(\Rightarrow \left[\begin{matrix} x-1=0\\ x^2+x-1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=1=y\\ x=\frac{-1\pm \sqrt{5}}{2}=y\end{matrix}\right.\)

Nếu $x^2+xy+y^2+2=0$

$\Leftrightarrow (x+\frac{y}{2})^2=-(2+\frac{3}{4}y^2)< 0$ (vô lý- loại)

Vậy............