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\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{2015}:2\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2013}{4030}\)
\(\frac{1}{x+1}=\frac{1}{2015}\)
=>x+1=2015
=>x=2014
\(\frac{x-10}{30}+\frac{x-14}{43}+\frac{x-5}{95}+\frac{x-148}{8}=0\)
\(\Rightarrow\frac{x-10}{30}-3+\frac{x-14}{43}-2+\frac{x-5}{95}-1+\frac{x-148}{8}+6=0\)
\(\Rightarrow\frac{x-100}{30}+\frac{x-100}{43}+\frac{x-100}{95}+\frac{x-100}{8}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{8}\right)=0\)
mà \(\frac{1}{30}+\frac{1}{43}+\frac{1}{95}+\frac{1}{8}\ne0\)
=> x - 100 = 0 => x = 100
Vậy x = 100
\(25\%.x-\frac{1}{5}.x=\frac{-1}{20}\)
\(=>\frac{1}{4}.x-\frac{1}{5}.x=\frac{-1}{20}\)
\(=>x.\left(\frac{1}{4}-\frac{1}{5}\right)=\frac{-1}{20}\)
\(=>x.\frac{1}{20}=\frac{-1}{20}\)
\(=>x=\frac{-1}{20}:\frac{1}{20}\)
\(=>x=-1\)
a)\(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\frac{45}{4}-\left(\frac{19}{7}+\frac{21}{4}\right)\)
\(=\frac{45}{4}-\left(\frac{76}{28}+\frac{147}{28}\right)\)
\(=\frac{45}{4}-\frac{223}{28}\)
\(=\frac{315}{28}-\frac{223}{28}\)
\(=\frac{23}{7}\)
b) \(\left(8\frac{5}{11}+3\frac{5}{8}\right)-3\frac{5}{11}\)
\(=\left(\frac{93}{11}+\frac{29}{8}\right)-\frac{38}{11}\)
\(=\left(\frac{744}{88}+\frac{319}{88}\right)-\frac{38}{11}\)
\(=\frac{1063}{88}-\frac{38}{11}=\frac{1063}{88}-\frac{304}{88}\)
\(=\frac{69}{8}\)
Ta có: \(\left(\frac{3x}{7}+1\right):\left(-4\right)=\frac{-1}{7}\)
\(\Rightarrow\)\(\frac{3x}{7}+1=\frac{-1}{7}.\left(-4\right)\)
\(\Rightarrow\)\(\frac{3x}{7}+1=\frac{4}{7}\)\(\Rightarrow\frac{3x}{7}=\frac{-3}{7}\)
\(\Rightarrow3x=-3\)\(\Rightarrow\)\(x=-1\)
Vậy x=-1
\(=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(=49\frac{8}{23}-14\frac{8}{23}-5\frac{7}{12}\)
\(=35-5\frac{7}{32}\)
\(=35-\frac{167}{32}\)
\(=\frac{1120}{32}-\frac{167}{32}\)
\(=\frac{935}{32}\)
\(49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(=35-5\frac{7}{32}\)
\(=35-\frac{167}{32}\)
\(=\frac{953}{32}\)