\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)

\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)

\(\left(x+3\right)\cdot\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)

mà \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\)

=> x + 3 = 0

x = -3

\(\Rightarrow\left(x+3\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\right)=0\\ \Rightarrow x=-3\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2010}+\dfrac{1}{2009}\ne0\right)\)

\(\dfrac{x+3}{2007}-\dfrac{x+3}{2008}=\dfrac{x+3}{2010}-\dfrac{x+3}{2009}\)

\(\Leftrightarrow x+3=0\)

hay x=-3

\(\frac{x+1}{2013}+\frac{x}{2012}+\frac{x-1}{2011}=\frac{x-2}{2010}+\frac{x-3}{2009}+\frac{x-4}{2008}\)

\(\Leftrightarrow\frac{x+1}{2013}-1+\frac{x}{2012}-1+\frac{x-1}{2011}-1=\frac{x-2}{2010}-1+\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)

\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}=\frac{x-2012}{2010}+\frac{x-2012}{2009}+\frac{x-2012}{2008}\)

\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}-\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(\Leftrightarrow x-2012=0\). Do \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

\(\Leftrightarrow x=2012\)

Ta có :\(A=3+3^2+3^3+...+3^{2008}\)(1)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2009}\)(2)

Lấy (2) trừ đi 1 ta có :

\(\Rightarrow2A=3^{2009}-3\)

Ta lại có :

\(2A+3=3^x\)

\(\Rightarrow3^{2009}=3^x\)

\(\Rightarrow x=2009\)

\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)

\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\) ( trừ 2 số bằng nhau)

\(\Rightarrow\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)

Mà \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\)

\(\Rightarrow x+3=0\)

\(\Rightarrow x=0-3=-3\)

Ta có:

\(\frac{x+4}{2008}+1+\frac{x+3}{2009}+1=\frac{x+2}{2010}+1+\frac{x+1}{2011}+1\)

\(\frac{x+2012}{2008}+\frac{x+2012}{2009}=\frac{x+2012}{2010}+\frac{x+2012}{2011}\)

\(\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

\(x=-2012\)

;Ko tồn tại nghiệm số thực OK

Ta có: \(A=3+3^2+3^3+...+3^{2008}\)

\(\Rightarrow3A=3^2+3^3+3^4+...+3^{2009}\)

Trừ \(3A-A=3^2+3^3+3^4+...+3^{2009}-3-3^2-3^3-...-3^{2008}\)

\(\Rightarrow2A=3^{2009}-3\)

Mà \(2A=3^x-3\)

\(\Rightarrow3^x=3^{2009}\)

\(\Rightarrow x=2009.\)

Vậy x = 2009.

\(a=3+3^2+3^3+...+3^{2008}\)

\(3a=3^2+3^3+3^4+...+3^{2009}\)

\(3a-a=\left(3^2+3^3+3^4+...+3^{2009}\right)-\left(3+3^2+3^3+...+3^{2008}\right)\)

\(2a=3^{2009}-3\)

\(2a+3=3^{2009}=3^x\)

\(x=2009\)

1/\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)

\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

Vì\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)nên để biểu thức có giá trị là 0 thì x+1=0 <=> x=-1

2/Tương tự bài 2 bạn cộng mỗi vế cho 3 mỗi biểu thức cộng cho 1 thỳ bn sẽ tìm đc kq là -2010

3/ trừ 2 cho mỗi vế, mỗi biểu thức trừ cho 1, lập luận ta có x=100

4/ bài này chuyển -3 qua vế trái thành 3 rồi tách, nhóm mỗi biểu thức với 1 ta có x=-10

1/�+12+�+13+�+14=�+15+�+162x+1​+3x+1​+4x+1​=5x+1​+6x+1​

⇔�+12+�+13+�+14−�+15−�+16=0⇔2x+1​+3x+1​+4x+1​−5x+1​−6x+1​=0

⇔(�+1)(12+13+14−15−16)=0⇔(x+1)(21​+31​+4