Sửa đề:

\(VP=\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}+\dfrac{2005}{2006}\)

Ta có: \(2005^2+1=\left(2005+1\right)^2-2.2005.1=2006^2-2.2005\)

\(\Rightarrow VP=\sqrt{2006^2-2.2005+\dfrac{2005^2}{2006^2}}+\dfrac{2005}{2006}\)

\(=\sqrt{\left(2006-\dfrac{2005}{2006}\right)^2}+\dfrac{2005}{2006}\)

\(=2006-\dfrac{2005}{2006}+\dfrac{2005}{2006}=2006\)

Phương trình đã cho tương đương

\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=2006\)

\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2006\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2006\)

Đến đây thì tự xét trường hợp và giải tìm nghiệm, bài này không cần điều kiện nhé

\(\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}=\dfrac{1}{2006}\sqrt{2006^2+2005^2+\left(2005.2006\right)^2}\)

\(=\dfrac{1}{2006}\sqrt{\left(2006-2005\right)^2+2.2005.2006+\left(2005.2006\right)^2}\)

\(=\dfrac{1}{2006}\sqrt{1+2.2005.2006+\left(2005.2006\right)^2}\)

\(=\dfrac{1}{2006}\sqrt{\left(2005.2006+1\right)^2}=\dfrac{2005.2006+1}{2006}=2005+\dfrac{1}{2006}\)

Phương trình tương đương:

\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2005+\dfrac{1}{2006}+\dfrac{2005}{2006}\)

\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2006\)

TH1: \(x\ge2\): \(x-1+x-2=2006\Rightarrow2x=2009\Rightarrow x=\dfrac{2009}{2}\)

TH2: \(x\le1\) : \(1-x+2-x=2006\Rightarrow-2x=2003\Rightarrow x=\dfrac{-2003}{2}\)

TH3: \(1< x< 2:\) \(x-1+2-x=2006\Rightarrow3=2006\) (vô nghiệm)

Vậy \(\left[{}\begin{matrix}x=\dfrac{2009}{2}\\x=\dfrac{-2003}{2}\end{matrix}\right.\)

\(\dfrac{x+4}{2016}+\dfrac{x+2}{2018}\ge\dfrac{x+14}{2006}+\dfrac{x+83}{1937}\)

\(\Leftrightarrow\dfrac{x+4}{2016}+1+\dfrac{x+2}{2018}+1\ge\dfrac{x+14}{2006}+1+\dfrac{x+83}{1937}+1\)

\(\Leftrightarrow\dfrac{x+2020}{2016}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{2006}-\dfrac{x+2020}{1937}\ge0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2016}+\dfrac{1}{2018}-\dfrac{1}{2006}-\dfrac{1}{1937}\right)\ge0\)

\(\Leftrightarrow x+2020\ge0\Leftrightarrow x\ge-2020\)

Vậy \(x\ge-2020\)

\(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)

⇔\(\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)

⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)

⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)

⇔ \(\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)

⇔ x+2009=0

⇔ x=-2009

vậy x=-2009 là nghiệm của pt

a) ( x² + x )² + 4( x² + x ) = 12

<=> ( x² + x )² + 4( x² + x ) + 4 - 16 = 0

<=> ( x² + x + 2)² - 16 = 0

<=> ( x² + x + 2 + 4)( x² + x + 2 - 4) = 0

<=> ( x² + x + 6 )( x² + x - 2) = 0

Do : x² + x + 6

= x² + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}+6-\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\) ≥ \(\dfrac{23}{4}\) > 0 ∀x

=> x² + x - 2 = 0

<=> x² - x + 2x - 2 = 0

<=> x( x - 1) + 2( x - 1) = 0

<=> ( x - 1)( x + 2 ) = 0

<=> x = 1 hoặc : x = - 2

KL.....

b) Kuroba kaito làm rùi nhé

\(\Leftrightarrow\dfrac{x}{2005}+1+\dfrac{x-1}{2006}+1=\dfrac{x-2}{2007}+1-1+1\)

\(\Leftrightarrow\dfrac{x+2005}{2005}+\dfrac{x+2005}{2006}=\dfrac{x+2005}{2007}\)

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}\right)=0\)

\(\Leftrightarrow x+2005=0\) (vì \(\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}\ne0\))

\(\Leftrightarrow x=-2005\)

\(\dfrac{x}{2005}+\dfrac{x-1}{2006}=\dfrac{x-2}{2007}-1\)

\(\Leftrightarrow\dfrac{x+2005}{2005}+\dfrac{x+2005}{2006}-\dfrac{x+2005}{2007}=0\)

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}\right)=0\)

\(\Leftrightarrow x=-2005\).

\(\dfrac{x}{2005}+\dfrac{x-1}{2006}=\dfrac{x-2}{2007}-1\\ \dfrac{x}{2005}+1+\dfrac{x-1}{2006}+1=\dfrac{x-2}{2007}+1\\ \dfrac{x+2005}{2005}+\dfrac{x+2005}{2006}=\dfrac{x+2005}{2007}\\ \left(x+2005\right)\left(\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}\right)=0\\ =>x+2005=0\\ =>x=-2005\)

Chúc bạn học tốt ^^

\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}=-4\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+4=0\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+1+1+1+1=0\\ \Rightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2005}+\dfrac{x+2010}{2006}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2005}+\dfrac{1}{2006}\right)=0\)

mà `1/2008+1/2007+1/2005+1/2006≠ 0`

`=> x+2010=0`

`=>x=-2010`

\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\)

=>x+2010=0

=>x=-2010

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

số số hạng của A là :

( 2007 - 3 ) : 3 + 1 = 669 ( số )

tổng A là :

( 2007 + 3 ) . 669 : 2 = 672345

B = \(\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)

B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\left(\dfrac{2005}{2}+1\right)+\left(\dfrac{2004}{3}+1\right)+...+\left(\dfrac{1}{2006}+1\right)+1}\)

B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2}+\dfrac{2007}{3}+...+\dfrac{2007}{2006}+\dfrac{2007}{2007}}\)

B = \(\dfrac{2006.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2007}\right)}{2007.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2006}+\dfrac{1}{2007}\right)}\)

B = \(\dfrac{2006}{2007}\)

\(A=3+6+9+12+...+2007\)

\(A=\dfrac{\left(2007+3\right)\left(\dfrac{2007-3}{3}+1\right)}{2}\)

\(A=\dfrac{2010.669}{2}=672235\)

\(B=\dfrac{\dfrac{2006}{2}+\dfrac{2006}{3}+\dfrac{2006}{4}+...+\dfrac{2006}{2007}}{\dfrac{2006}{1}+\dfrac{2005}{2}+\dfrac{2004}{3}+...+\dfrac{1}{2006}}\)

\(B=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{2006+\left(\dfrac{2005}{2}+1\right)+\left(\dfrac{2004}{3}+1\right)+...\left(\dfrac{1}{2006}+1\right)-2005}\)

\(B=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{\dfrac{2007}{2007}+\dfrac{2007}{2}+\dfrac{2005}{3}+...+\dfrac{2007}{2006}}\)

\(B=\dfrac{2006\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2007}\right)}{2007\left(\dfrac{1}{2007}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}\right)}=\dfrac{2006}{2007}\)

tik mik nha !!!

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