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

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 

(1): \(\frac{x - 2013}{2011}+\frac{x - 2011}{2009}=\frac{x - 2009}{2007}+\frac{x - 2007}{2005}\)

=>\(\left(\frac{x - 2013}{2011} + 1\right) + \left(\frac{x - 2011}{2009} + 1\right) = \left(\frac{x - 2009}{2007} + 1\right) + \left(\frac{x - 2007}{2005} + 1\right)\)

=>\(\frac{x - 2}{2011}+\frac{x - 2}{2009}=\frac{x - 2}{2007}+\frac{x - 2}{2005}\)

=>\((x-2)\left(\frac{1}{2011}+\frac{1}{2009}-\frac{1}{2007}-\frac{1}{2005}\right)=0\)

=>x-2=0

=>x=2

(2): ĐKXĐ: x<>1

\(\frac{x^2 - (2 - m)x - 2m}{x - 1}=0\)

=>\(\frac{(x - 2)(x + m)}{x - 1}=0\quad(x<>1)\)

=>(x-2)(x+m)=0

=>x=2 hoặc x=-m

Để (1) và (2) tương đương thì (2) có nghiệm duy nhất là x=2

TH1: -m=2

=>m=-2

(2) sẽ trở thành: \(\frac{\left(x-2\right)\left(x-2\right)}{x-1}=0\)

=>\(\frac{\left(x-2\right)^2}{x-1}=0\)

=>\(\left(x-2\right)^2=0\)

=>x-2=0

=>x=2(nhận)

TH2: x=-m bị loại do không thỏa điều kiện x<>1

=>-m=1

=>m=-1

(2) sẽ trở thành: \(\frac{(x - 2)(x - 1)}{x - 1}=0\)

=>x-2=0

=>x=2

=>Nhận

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....

Ta có: \(\frac{x-2013}{2011}+\frac{x-2011}{2009}=\frac{x-2009}{2007}+\frac{x-2007}{2005}\)

=>\(\frac{x-2013}{2011}+1+\frac{x-2011}{2009}+1=\frac{x-2009}{2007}+1+\frac{x-2007}{2005}+1\)

=>\(\frac{x-2}{2011}+\frac{x-2}{2009}=\frac{x-2}{2007}+\frac{x-2}{2005}\)

=>x-2=0

=>x=2

Thay x=2 vào (2), ta được:

\(\frac{2^2-\left(2-m\right)\cdot2-2m}{2-1}=0\)

=>4-2m-2(2-m)=0

=>4-2m-4+2m=0

=>0m=0(luôn đúng)

=>m∈R