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\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)

=>x+2011=0

hay x=-2011

28 tháng 4 2018

\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)

\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)

\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)

\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)

\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)

28 tháng 4 2018

x=-2011

1 tháng 2 2017

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

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

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

\(\Leftrightarrow x-6027=0\Leftrightarrow x=6027\)

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)

\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)

=>x-2014=0

hay x=2014

a: =>3,6-x+0,5=3,5-0,75+x

=>4,1-x=x+2,75

=>-2x=-1,35

=>x=0,675

b: =>5x^2-5x+x-1=0

=>(x-1)(5x+1)=0

=>x=1 hoặc x=-1/5

c: \(\Leftrightarrow\left(\dfrac{2-x}{2008}+1\right)=\left(\dfrac{1-x}{2009}+1\right)+\left(1-\dfrac{x}{2010}\right)\)

=>\(2010-x=0\)

=>x=2010

20 tháng 2 2023

Cảm ơn ạ thank you

 

20 tháng 2 2019

<=>2-1-1 = \(\frac{x}{2008}\)\(\frac{x}{2009}\)\(\frac{x}{2010}\)

<=>0= \(\frac{x}{2008}\)\(\frac{x}{2009}\) - \(\frac{x}{2010}\)

=>x=0

29 tháng 3 2022

refer

Hỏi đáp Toán

Giải bài 18 trang 14 SGK Toán 8 Tập 2 | Giải toán lớp 8

2 tháng 3 2022

`Answer:`

\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)

\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)

\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)

\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)

\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(\Rightarrow x+2014=0\)

\(\Leftrightarrow x=-2014\)