от ammornil » 24 Ное 2012, 20:56
[tex](x+1)^4=2.(1+x^4)+3.x^2[/tex]
[tex](x+1)^4=2.x^4+3.x^2+2 \hspace{8} \Leftrightarrow \hspace{8} x^4+4.x^3+6.x^2+4.x+1= 2.x^4+3.x^2+2 \hspace{8} \Rightarrow \\
x^4-4.x^3-3.x^2-4.x+1=0 \\
x^4-4.x^3+6.x^2-4.x+1-9.x^2=0 \\
(x-1)^4-(3.x)^2=0 \hspace{16} \Leftrightarrow \hspace{8} ((x-1)^2-3.x).((x-1)^2+3.x)=0 \\
(x^2-2.x+1-3.x).(x^2-2.x+1+3.x)=0 \hspace{8} \Leftrightarrow \hspace{8} (x^2-5.x+1).(x^2+x+1)=0 \Rightarrow \\
x^2-5.x+1=0 \hspace{8} \cup \hspace{8} x^2+x+1=0 \\
x_{_{1,2}}=\frac{5 \pm \sqrt{21}}{2} \hspace{16} \cup \hspace{16} \oslash \\
x_1=\frac{5-\sqrt{21}}{2} \hspace{8} x_2=\frac{5+\sqrt{21}}{2}[/tex]
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]