chào chào ..... hotgirl quay lại mang toán hay đến nè
giải hpt bậc 3 nhá
[tex]\left{\begin{2x+x^2y=y}\\{2y+y^2z=z}\\{2z+z^2x=x}[/tex]
từ từ mà làm nhá rùa ;
[tex]\left{\begin{2x+x^2y=y}\\{2y+y^2z=z}\\{2z+z^2x=x}[/tex]
[tex] \Leftrightarrow \left{\begin{y(1-x^2)=2x}\\{z(1-y^2)=2y}\\{x(1-z^2)=2z}[/tex]
[tex]\Leftrightarrow \left{\begin{x=\frac{2z}{1-z^2}}\\{y=\frac{2x}{1-x^2}}\\{z=\frac{2y}{1-y^2}}(DK:x,y,z \neq \pm \ 1)[/tex]
Đặt [tex]x=tanv(1) \Rightarrow y=\frac{2.tanv}{1-tan^2v}=\frac{2.tanv}{\frac{2tanv}{tan2v}}=tan2v[/tex]
[tex]\Rightarrow z=\frac{2.tan2v}{1-tan^22v}=\frac{2tan2v}{\frac{2tan2v}{tan4v}}=tan 4v [/tex]
[tex] x=\frac{2.tan4z}{1-tan^24v}=\frac{2tan4z}{\frac{2tan4z}{tan8v}}=tan8v(2)[/tex]
[tex](1) &(2) \Rightarrow tan8v=tanv \Rightarrow v=\frac{k.\pi}{7} k = (1\leq k \leq 6; k \in\ Z)[/tex]
[tex]\Rightarrow \left{\begin{x=tanv=tan\frac{k.\pi}{7}}\\{y=tan2v=tan2\frac{k.\pi}{7}}\\{z=tan4v=tan\frac{4k.\pi}{7}}[/tex]
