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8 tháng 7 2017
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
T
22 tháng 10 2018
\(\left(3x+1\right)^2=25\)
\(\Rightarrow\left(3x+1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow\orbr{\begin{cases}3x+1=5\\3x+1=-5\end{cases}\Rightarrow\orbr{\begin{cases}3x=5-1=4\\3x=-5-1=-6\end{cases}}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-2\end{cases}}\)
\(\left[x-\frac{1}{2}\right]+\frac{1}{2}=\frac{5}{8}\)
\(\Rightarrow x-0=\frac{5}{8}\)
\(x=\frac{5}{8}\)
\(\left[x+\frac{3}{4}\right]-\frac{1}{3}=0\)
\(x+\frac{3}{4}=0+\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}-\frac{3}{4}\)
\(x=\frac{-5}{12}\)
IL
1
LM
7 tháng 9 2016
a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)
b.
\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)
\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)
\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)
\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)
PD
2
IM
29 tháng 8 2016
\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)
\(\Rightarrow x.0+1=0\)
=> 1=0 ( Vô lý )
Vậy \(x\in\varnothing\)
29 tháng 8 2016
\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow x.\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)=1\)
\(\Rightarrow x.0=1\Rightarrow x=0\)
Vậy x=0