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\(\Rightarrow\)\(\frac{2}{6}\)+ \(\frac{2}{12}\)+ \(\frac{2}{20}\)+...+\(\frac{2}{x\left(x+1\right)}\)= \(\frac{2011}{2013}\)
\(\Rightarrow\)\(\frac{2}{2.3}\)+ \(\frac{2}{3.4}\)+ \(\frac{2}{4.5}\)+...+ \(\frac{2}{x\left(x+1\right)}\)= \(\frac{2011}{2013}\)
\(\Rightarrow\)\(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+...+ \(\frac{1}{x}\)- \(\frac{1}{x+1}\)= \(\frac{2011}{2013}\): 2
\(\Rightarrow\)\(\frac{1}{2}\)- \(\frac{1}{x+1}\)= \(\frac{2011}{4026}\)
\(\Rightarrow\)\(\frac{1}{x+1}\)= \(\frac{1}{2}\)- \(\frac{2011}{4026}\)= \(\frac{1}{2013}\)
\(\Rightarrow\)\(x+1=2013\)
a) \(x.2\frac{1}{4}=2\frac{3}{8}-\frac{3}{4}\Rightarrow x.\frac{9}{4}=\frac{19}{8}+\frac{-3}{4}\)\(\Rightarrow x.\frac{9}{4}=\frac{13}{8}\Rightarrow x=\frac{13}{8}:\frac{9}{4}\Rightarrow x=\frac{13}{18}\)
Vậy \(x=\frac{13}{18}\)
b) \(\frac{1}{3}:2.x-\frac{4}{5}=\frac{2}{3}\Rightarrow\frac{1}{3}:2.x=\frac{2}{3}+\frac{4}{5}\)\(\Rightarrow\frac{1}{3}:2.x=\frac{22}{15}\Rightarrow2.x=\frac{1}{3}:\frac{22}{15}\Rightarrow2.x=\frac{5}{22}\)\(\Rightarrow x=\frac{5}{22}.2\Rightarrow x=\frac{5}{44}\)
Vậy \(x=\frac{5}{44}\)
c) \(\frac{1}{3}x:10\frac{1}{5}=\frac{5}{3}\Rightarrow\frac{1}{3}x:\frac{51}{5}=\frac{5}{3}\)\(\Rightarrow\frac{1}{3}x=\frac{5}{3}.\frac{51}{5}\Rightarrow\frac{1}{3}x=17\Rightarrow x=17:\frac{1}{3}\Rightarrow x=51\)
Vậy \(x=51\)
d) \(5\frac{1}{7}:\left(x-\frac{1}{3}\right)=3\frac{1}{2}\Rightarrow\frac{36}{7}:\left(x-\frac{1}{3}\right)=\frac{7}{2}\)\(\Rightarrow x-\frac{1}{3}=\frac{36}{7}:\frac{7}{2}\Rightarrow x-\frac{1}{3}=\frac{72}{49}\)\(\Rightarrow x=\frac{72}{49}+\frac{1}{3}\Rightarrow x=\frac{265}{147}\)
Vậy \(x=\frac{265}{147}\)
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