1. Tìm x
a) \(2\frac{1}{3}\)+ ( x - \(\frac{3}{2}\)) = ( 3 - \(\frac{3}{2}\)) x
b) x + ( x + 1 ) + ( x + 2 ) + .... + ( x + 2018) = 2019
c) \(\frac{1}{3}\)+ x = \(\frac{1}{12}\)+ \(\frac{1}{20}\)+ \(\frac{1}{30}\)+ \(\frac{1}{42}\)+ \(\frac{1}{56}\)+ \(\frac{1}{72}\)+ \(\frac{1}{90}\)+ \(\frac{1}{110}\)
d) x + \(\frac{1}{13}\)- \(\frac{1}{15}\)- \(\frac{1}{35}\)- \(\frac{1}{63}\)- \(\frac{1}{99}\)- \(\frac{1}{143}\)= 2x + \(\frac{1}{2}\)
Giúp em vs!!!!!! >.<
Bài 1c)
\(\frac{1}{3}+x=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\)
\(\frac{1}{3}+x=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)
\(\frac{1}{3}+x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{3}+x=\frac{1}{3}-\frac{1}{11}=\frac{11}{33}-\frac{3}{33}=\frac{8}{33}\)
\(x=\frac{8}{33}-\frac{1}{3}=\frac{8}{33}-\frac{11}{33}=\frac{-3}{33}=\frac{-1}{11}\)
