giải hệ nữa nhé!
$\left\{\begin{matrix} x-2\sqrt{3y-2}=2y-3\sqrt[3]{2x-1} & \\ 2y-2-2\sqrt{3y-2}=x-\sqrt[3]{2x-1} & \end{matrix}\right.$
Lời giải. Điều kiện $y \ge \frac 23$.
Từ phương trình thứ hai ta thu được $\sqrt[3]{2x-1}=x+2+2 \sqrt{3y-2}-2y$. Thay $\sqrt[3]{2x-1}$ vào phương trình đầu ta được $$x-2 \sqrt{3y-2}=2y-3(x+2+2 \sqrt{3y-2}-2y) \Leftrightarrow 2x+2\sqrt{3y-2}=4y-3$$
Ta có $x=2y- \sqrt{3y-2}- \frac 32$.
Thay $x$ vào phương trình thứ hai ta được $$2y-2-2 \sqrt{3y-2}= 2y- \sqrt{3y-2}- \frac 32 - \sqrt[3]{2x-1} \Leftrightarrow 2\sqrt{3y-2}+1=2\sqrt[3]{2x-1} \qquad (1)$$
Ta có $$\begin{aligned} (1) & \Leftrightarrow \left( 2 \sqrt{3y-2}+1 \right)^3= \left( 2 \sqrt[3]{2x-1} \right)^3 \\ & \Leftrightarrow 8 \sqrt{(3y-2)^3}+1+12(3y-2)+6\sqrt{3y-2}=8(2x-1) \qquad (2) \end{aligned}$$
Đặt $\sqrt{3y-2}=a \ge 0$ thì $3x=6y-3 \sqrt{3y-2}-\frac 92= 2(3y-2)-3 \sqrt{3y-2}- \frac 12=2a^2-3a- \frac 12$.
Do đó $$\begin{aligned}(2) & \Leftrightarrow 8a^3+1+12a^2+6a=8 \left( \frac 43 a^2-2a- \frac 13 -1 \right) \\ & \Leftrightarrow 24a^3+3+36a^2+18a=8(4a^2-6a-4) \\ & \Leftrightarrow 24a^3+4a^2+66a+35=0 \\ & \Leftrightarrow (2a+1)(12a^2-4a+35)=0 \end{aligned}$$
Ta tìm được $a= \frac{-1}{2}<0$, mâu thuẫn.
Vậy hệ vô nghiệm.
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).