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\(\frac{34000-68}{102000-204}=\frac{33932}{101796}=\frac{33932}{33932.3}=\frac{1}{3}\)
hay chung to rang cac phan so sau deu bang nhau:1945 -19/1999 va 194545-1945/19990
1/3 va 34000-68/102000-204
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}.\)
\(=2\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{192}\right)\)
\(=2\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{96}-\frac{1}{192}\right)\)
\(=2\times\left(1-\frac{1}{192}\right)\)
\(=2\times\frac{191}{192}=\frac{191}{68}\)
\(\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}\)
\(=\frac{1}{3.1}+\frac{1}{3.2}+\frac{1}{3.2^2}+...+\frac{1}{3.2^6}\)
\(=\frac{1}{3}.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)
\(=\frac{1}{3}.A\)với \(A=\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)
\(\Rightarrow2A=2.\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)
\(\Rightarrow2A=2+\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2^5}\)
\(\Rightarrow2A-A=\left(2+\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2^5}\right)-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)
\(\Rightarrow A=2-\frac{1}{2^6}=2-\frac{1}{64}=\frac{127}{64}\)
\(\Rightarrow\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+...+\frac{2}{192}=\frac{1}{3}.\frac{127}{64}=\frac{127}{192}\)
\(\dfrac{2}{5}\)= \(\dfrac{6}{15}\)> \(\dfrac{6}{16}\) = \(\dfrac{3}{8}\) > \(\dfrac{3}{9}\) = \(\dfrac{1}{3}\) = \(\dfrac{1\times5}{3\times5}\) = \(\dfrac{5}{15}\) > \(\dfrac{5}{16}\) vậy \(\dfrac{2}{5}\) > \(\dfrac{3}{8}\) > \(\dfrac{1}{3}\) > \(\dfrac{5}{16}\)
\(\dfrac{5}{16}\) = \(\dfrac{5\times4}{16\times4}\) = \(\dfrac{20}{64}\) > \(\dfrac{20}{65}\) = \(\dfrac{4}{13}\)
Các phân số đã cho được sắp xếp theo thứ tự tăng dần là:
\(\dfrac{4}{13}\); \(\dfrac{5}{16}\); \(\dfrac{1}{3}\); \(\dfrac{3}{8}\); \(\dfrac{2}{5}\)