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\(\frac{x-1}{13}+\frac{x-2}{12}+\frac{x-3}{11}-3=0\)

\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)+\left(\frac{x-2}{12}-1\right)+\left(\frac{x-3}{11}-1\right)\)

\(\Leftrightarrow\frac{x-14}{13}+\frac{x-14}{12}+\frac{x-14}{11}=0\)

\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)=0\). Vì \(\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)>0\)

\(\Leftrightarrow x-14=0\Rightarrow x=0+14=14\). Vậy \(x=14\)

26 tháng 8 2020

Bài làm:

Ta có: \(\frac{x-1}{13}+\frac{x-2}{12}+\frac{x-3}{11}-3=0\)

\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)+\left(\frac{x-2}{12}-1\right)+\left(\frac{x-3}{11}-1\right)=0\)

\(\Leftrightarrow\frac{x-14}{13}+\frac{x-14}{12}+\frac{x-14}{11}=0\)

\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)=0\)

\(\Leftrightarrow x-14=0\)

\(\Rightarrow x=14\)

21 tháng 7 2019

a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)

=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

=>  x + 1 = 0

=> x = -1

21 tháng 7 2019

b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)

=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)

=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)

=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)

=> x - 2021 = 0

=> x = 2021

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

=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)

=> \(-\frac{1}{12}x+6=7\)

=> \(-\frac{1}{12}x=1\)

=> x = -12

6 tháng 8 2016

a)

\(\Rightarrow3^x\left(3^2+3+1\right)=117\)

\(\Rightarrow3^x.13=117\)

\(\Rightarrow3^x=9\)

\(\Rightarrow3^x=3^2\)

=>x=2

b)

\(3^{2x+1}=3^{-4}\)

=> 2x+1= - 4

=>\(x=-\frac{5}{2}\)

c)

\(\left(x+2\right)^4=16\)

\(\Rightarrow\left[\begin{array}{nghiempt}\left(x+2\right)^4=2^4\\\left(x+2\right)^4=\left(-2\right)^4\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+2=2\\x+2=-2\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\-4\end{array}\right.\)

 

6 tháng 8 2016

thanks bn nhiu

haha

20 tháng 9 2018

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

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

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

\(\Rightarrow\) \(\left(x+2012\right).\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}\right)\)

Mà \(\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}\right)\ne0\)

\(\Rightarrow\) \(x+2012=0\)

\(\Rightarrow\) \(x=-2012\)

10 tháng 3 2016

thay x=1 tổng hệ số =1 nha nguyệt

3 tháng 2 2016

Bài toán khó đấy