tìm x biết :x phần 2 = x-1 phần 3
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\(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(\Rightarrow3x-\frac{3}{2}-5x-3=-x+\frac{1}{5}\)
\(\Rightarrow3x-5x+x=\frac{1}{5}+\frac{3}{2}+3\)
\(\Rightarrow-x=\frac{17}{10}+3\)
\(\Rightarrow-x=\frac{47}{10}\)
\(\Rightarrow x=-\frac{47}{10}\)

a) \(\frac{2}{3x}=\frac{5}{2}\Leftrightarrow3x=\frac{4}{5}\Rightarrow x=\frac{4}{15}\)
b) \(\frac{x-3}{4}=\frac{1}{2}\Rightarrow x-3=2\Rightarrow x=5\)
c) \(\frac{5}{24+x}=\frac{7}{12}\Rightarrow24+x=\frac{60}{7}\Rightarrow x=\frac{-108}{7}\)
d) \(-6x=18\Rightarrow x=-3\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.(x+1)}=\frac{2007}{2009}\)
=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2019}\)
Vì 1 = 1
=> x + 1 = 2019
=> x = 2019 - 1
=> x = 2018
\(\dfrac{x}{2}=\dfrac{x-1}{3}\)
\(\Leftrightarrow3x=2x-2\)
hay x=-2
\(\dfrac{x}{2}=\dfrac{x-1}{3}\\ \Rightarrow3x=2x-2\\ \Rightarrow x=-2\)