\(x^2-x-1=0\)

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 

a) \(0,25x^3+x^2+x=0\)

\(\Leftrightarrow x\left(0,25x^2+x+1\right)=0\)

\(\Leftrightarrow x\left[\left(\frac{1}{2}x\right)^2+2\cdot\frac{1}{2}x\cdot1+1^2\right]=0\)

\(\Leftrightarrow x\left(\frac{1}{2}x+1\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

Vậy....

b) \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)

\(\Leftrightarrow\frac{2-x}{2007}-1+2=\frac{1-x}{2008}+1+\frac{-x}{2009}+1\)

\(\Leftrightarrow\frac{2-x+2007}{2007}=\frac{1-x+2008}{2008}+\frac{-x+2009}{2009}\)

\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}+\frac{2009-x}{2009}\)

\(\Leftrightarrow\frac{2009-x}{2007}-\frac{2009-x}{2008}-\frac{2009-x}{2009}=0\)

\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

Vì \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)

\(\Rightarrow2009-x=0\)

\(\Leftrightarrow x=2009\)

Vậy....