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bạn quy đồng mẫu số tất cả cả các phân số trên lên thành 243 rồi cộng vào là được
S = 1/3+1/9+1/27+1/81+1/243+1/729+1/2187 ( 1 )
Nhân S với 3. Ta có:
S x 3 = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729 ( 2 )
Trừ ( 2 ) với ( 1 ) ta có:
S x 3 - S = 1 - 1/ 2187
2S = 2186/ 2187
S = 2186/ 2187 : 2
S = 1093/ 2187
đặt S=\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
=>3S= \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
=>3S-S=\(\left(1+\frac{1}{3}+...+\frac{1}{243}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}\right)\)
=>s=1-1/729 = 728/729
1/3+1/9+1/27+1/81+1/243+1/729=(1/3+1/9+1/81)+(1/27+1/243+1/729)=37/81+37/729=333/729+37/729=370/729
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(=\frac{3}{9}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(=\frac{4}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(=\frac{12}{27}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{13}{27}+\frac{1}{81}+\frac{1}{243}=\frac{39}{81}+\frac{1}{81}+\frac{1}{243}=\frac{40}{81}+\frac{1}{243}\)
\(=\frac{120}{243}+\frac{1}{243}=\frac{121}{243}\)
1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
=1+ 243/729+ 81/729 + 27/729 + 9/729 + 3/729
=1093/729
Ta có:\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
Xét\(\frac{1}{3}A=\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Leftrightarrow A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{729}\)
\(\Leftrightarrow\frac{2}{3}A=\frac{243-1}{729}\Leftrightarrow A=\frac{3}{2}\times\frac{242}{729}=\frac{121}{243}\)
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
1+ 1 /3+1/9+1/27+1/81+1/243+1/729.
Đặt:
S = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
Nhân S với 3 ta có:
S x 3 = 3 +1+ 1/3 + 1/9 + 1/27 + 1/81
Vậy:
S x 3 - S = 3 - 1/243
2S = 728/243
S = 364/243
tick đúng nha
Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\) ta có :
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)\)
\(2A=1-\frac{1}{3^5}\)
\(A=\frac{1-\frac{1}{3^5}}{2}\)
Vậy \(A=\frac{1-\frac{1}{3^5}}{2}\)
Chúc bạn học tốt ~
Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A x 3 = 3 x (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243
A x 3 - A = 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - (1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/3 + 1/9 + 1/27 + 1/81 + 1/243 - 1/3 - 1/9 - 1/27 - 1/81 - 1/243 - 1/729
= 1 - 1/729
A x 2 = 728/729
A = 364/729
là121/243