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1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)
=> \(\dfrac{2}{2}\) + \(\dfrac{2}{6}\) + \(\dfrac{2}{12}\) + \(\dfrac{2}{20}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)
=> \(\dfrac{2}{1.2}\) + \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)
=> \(\dfrac{1.2}{1.2}\) + \(\dfrac{1.2}{2.3}\) + \(\dfrac{1.2}{3.4}\) + \(\dfrac{1.2}{4.5}\) + ... + \(\dfrac{1.2}{x\left(x+1\right)}\) = 1\(\dfrac{2015}{2017}\)
=> 2(\(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{x\left(x+1\right)}\)) = 1\(\dfrac{2015}{2017}\)
=> 2(1 - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\)) = 1\(\dfrac{2015}{2017}\)
=> 2(1 - \(\dfrac{1}{x+1}\)) = \(\dfrac{4032}{2017}\)
=> 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{4032}{2017}\) : 2
=> 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{2016}{2017}\)
=> \(\dfrac{1}{x+1}\) = 1 - \(\dfrac{2016}{2017}\)
=> \(\dfrac{1}{x+1}\) = \(\dfrac{1}{2017}\)
=> x + 1 = 2017
=> x = 2017 - 1
=> x = 2016
a) Vì \(\dfrac{x+5}{3}\)= \(\dfrac{x-6}{7}\) nên 7(x+5) = 3(x-6)
=> 7x+ 35 = 3x - 18
7x - 3x = -18 -35
4x = -53
x = -53:4
x = \(\dfrac{-53}{4}\)
\(\dfrac{1}{3}+\dfrac{1}{6}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2017}{2019}\\ \Rightarrow\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2017}{2019}\\ \Rightarrow2.\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2017}{2019}\\ \Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2017}{4038}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2019}\\ \Rightarrow x=2018\)
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
\(\Leftrightarrow2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2015}{2017}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2015}{4034}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2015}{4034}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2017}\)
=>x+1=2017
hay x=2016