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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
Dấu "." là dấu nhân bạn nhé.
Ta có:
\(A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+...+\dfrac{1}{729.3}\)
\(\Rightarrow3A=3+1+\dfrac{1}{3}+\dfrac{1}{9}+...+\dfrac{1}{729}\)
\(\Rightarrow3A-A=3-\dfrac{1}{729.3}\)
\(\Rightarrow2A=3-\dfrac{1}{729.3}\)
\(\Rightarrow A=\dfrac{1}{2}\left(3-\dfrac{1}{729.3}\right)\)
A= 1/3 + 1/9 + 1/27 + 1/81 + 1/243
Ax3=(1/3 + 1/9 + 1/27 + 1/81 + 1/243)x3
Ax3=1/3 x 3 + 1/9 x 3 + 1/27 x 3 + 1/81 x 3 + 1/243 x 3
Ax3=1+1/3+1/9+1/27+1/81
Ax3-A=(1+1/3+1/9+1/27+1/81)-(1/3+1/9+1/27+1/81+1/243)
Ax(3-1)=1-1/243
Ax2=243/243-1/243
Ax2= 242/243
A = 242/243:2
A = 242/243 x 1/2
A = 121/243
Vậy A= 121/243
Li-ke cho mik nhé!
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Lời giải:
Đặt $A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}$
$3\times A=3+1+\frac{1}{3}+\frac{1}{9}+...+\frac{1}{729}$
$3\times A-A=3-\frac{1}{2187}$
$2\times A=3-\frac{1}{2187}=\frac{6560}{2187}$
$A=\frac{6560}{2187}:2=\frac{3280}{2187}$