Question

Tính giá trị biểu thức: A= 1/3+1/9+1/27+1/81+1/243

Answer 1

Ta có:\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

Xét\(\frac{1}{3}A=\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)

\(\Leftrightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{729}\)

\(\Leftrightarrow\frac{2}{3}A=\frac{243-1}{729}\Leftrightarrow A=\frac{3}{2}\times\frac{242}{729}=\frac{121}{243}\)

Answer 2

Phải là : A=1/3+1/9+1/27+1/81+1/243 ta có: 3A=1+1/3+1/9+1/27+1/81                     3A-A=(1+1/3+1/9+1/27+1/81)-(1/3+1/9+1/27+1/81+1/243)=1-1/243       2A=242/243                                             A=242/243:2=121/243

Answer 3

A×3=3×(1/3+1/9+1/27+1/81+1/243) A×3=1+1/3+1/9+1/27+1/81 A×3-A=(1+1/3+1/9+1/27+1/81)-(1/3+1/9+1/27+1/81+1/243 A×2=1-1/243 A×2=242/243 A=242/243:2 A= 121/243