\(x^2-x-1=0\)
Ta có \(\Delta=b^2-4ac=\left(-1\right)^2-4.1.\left(-1\right)=1+4=5>0\); \(\sqrt{\Delta}=\sqrt{5}\)
Phuông trình có 2 nghiệm phân biệt
\(a=x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{1+\sqrt{5}}{2}\)
\(b=x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{1-\sqrt{5}}{2}\)
Ta có \(a^{2007}+b^{2007}+a^{2009}+b^{2009}\)
\(\Leftrightarrow a^{2007}.\left(1+a^2\right)+b^{2007}.\left(1+b^2\right)\)
\(\Leftrightarrow\left(\frac{1+\sqrt{5}}{2}\right)^{2007}.\left(1+\left(\frac{1+\sqrt{5}}{2}\right)^2\right)+\left(\frac{1-\sqrt{5}}{2}\right)^{2007}.\left(1+\left(\frac{1-\sqrt{5}}{2}\right)^2\right)\)
\(\Leftrightarrow\left(\frac{1+\sqrt{5}}{2}\right)^{2007}.\left(1+\frac{3+\sqrt{5}}{2}\right)+\left(\frac{1-\sqrt{5}}{2}\right)^{2007}.\left(1+\frac{3-\sqrt{5}}{2}\right)\)
\(\Leftrightarrow\left(\frac{1+\sqrt{5}}{2}\right)^{2007}.\left(\frac{5+\sqrt{5}}{2}\right)+\left(\frac{1-\sqrt{5}}{2}\right)^{2007}.\left(\frac{5-\sqrt{5}}{2}\right)\)
\(\Leftrightarrow\sqrt{5}.\left(\frac{1+\sqrt{5}}{2}\right)^{2008}+\sqrt{5}.\left(\frac{1-\sqrt{5}}{2}\right)^{2008}\)
\(\Leftrightarrow\sqrt{5}.\left[\left(\frac{1+\sqrt{5}}{2}\right)^{2008}+\left(\frac{1-\sqrt{5}}{2}\right)^{2008}\right]⋮5\) (ĐPCM)
Nhớ k cho mình nhé
