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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
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
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)
\(\left(x+2\right)^4=16\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x+2\right)^4=2^4\\\left(x+2\right)^4=\left(-2\right)^4\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+2=2\\x+2=-2\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\-4\end{array}\right.\)
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 \(\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}\right)\ne0\)
\(\Rightarrow\) \(x+2012=0\)
\(\Rightarrow\) \(x=-2012\)
\(\frac{x-1}{13}+\frac{x-2}{12}+\frac{x-3}{11}-3=0\)
\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)+\left(\frac{x-2}{12}-1\right)+\left(\frac{x-3}{11}-1\right)\)
\(\Leftrightarrow\frac{x-14}{13}+\frac{x-14}{12}+\frac{x-14}{11}=0\)
\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)=0\). Vì \(\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)>0\)
\(\Leftrightarrow x-14=0\Rightarrow x=0+14=14\). Vậy \(x=14\)
Bài làm:
Ta có: \(\frac{x-1}{13}+\frac{x-2}{12}+\frac{x-3}{11}-3=0\)
\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)+\left(\frac{x-2}{12}-1\right)+\left(\frac{x-3}{11}-1\right)=0\)
\(\Leftrightarrow\frac{x-14}{13}+\frac{x-14}{12}+\frac{x-14}{11}=0\)
\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)=0\)
\(\Leftrightarrow x-14=0\)
\(\Rightarrow x=14\)