bạn quy đồng mẫu số tất cả cả các phân số trên lên thành 243 rồi cộng vào là được
\(C=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(C=\frac{81}{243}+\frac{27}{243}+\frac{9}{243}+\frac{3}{243}+\frac{1}{243}\)
\(C=\frac{81+27+9+3+1}{243}=\frac{121}{243}\)
Vậy C = ...
Bài giải
\(C=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(C=\frac{1}{3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot3\cdot3}+\frac{1}{3\cdot3\cdot3\cdot3}+\frac{1}{3\cdot3\cdot3\cdot3\cdot3}\)
\(3C=1+\frac{1}{3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot3\cdot3} +\frac{1}{3\cdot3\cdot3\cdot3}\)
\(3C-C=1-\frac{1}{243}\)
\(2C=\frac{242}{243}\)
\(C=\frac{242}{243}\text{ : }2=\frac{121}{243}\)
Bài giải
Ta có :
\(C=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(C=\frac{1}{3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot3\cdot3}+\frac{1}{3\cdot3\cdot3\cdot3}+\frac{1}{3\cdot3\cdot3\cdot3\cdot3}\)
\(3C=1+\frac{1}{3}+\frac{1}{3\cdot3}+\frac{1}{3\cdot3\cdot3} +\frac{1}{3\cdot3\cdot3\cdot3}\)
\(3C-C=1-\frac{1}{3\cdot3\cdot3\cdot3\cdot3}=1-\frac{1}{243}\)
\(2C=\frac{242}{243}\)
\(C=\frac{242}{243}\text{ : }2=\frac{121}{243}\)
