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Đặt \(A=\frac{1}{3}+\frac{1}{9}+.......+\frac{1}{59049}\)
\(3A=3.\left(\frac{1}{3}+\frac{1}{9}+......+\frac{1}{59049}\right)\)
\(3A=1+\frac{1}{3}+........+\frac{1}{19683}\)
\(3A-A=\left(1+\frac{1}{3}+......+\frac{1}{19683}\right)-\left(\frac{1}{3}+\frac{1}{9}+........+\frac{1}{59049}\right)\)
\(2A=1-\frac{1}{59049}\)
\(2A=\frac{59048}{59049}\)
\(A=\frac{59048}{59049}:2\)
\(A=\frac{59048}{118098}\)
\(\text{Đặt : }A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=1-\frac{1}{729}\)
\(\Rightarrow2A=\frac{728}{729}\)
\(\Rightarrow A=\frac{728}{729}:2=\frac{364}{729}\)
Gọi tong trên là A
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}+\frac{1}{7129}+\frac{1}{2187}\)
\(3A=\frac{1}{3}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{729}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\right)\)
\(2A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}-\frac{1}{243}-\frac{1}{729}-\frac{1}{2187}\)
\(2A=1-\frac{1}{2187}\)
\(2A=\frac{2186}{2187}\)
\(A=\frac{2186}{2187}:2\)
\(A=\frac{1093}{2187}\)
Vậy tổng A = \(\frac{1093}{2187}\)
\(3y=3\cdot\frac{1}{1}+3\cdot\frac{1}{3}+3\cdot\frac{1}{9}+...+3\cdot\frac{1}{729}+3\cdot\frac{1}{2187}\)
\(=3+\frac{1}{1}+\frac{1}{3}...+\frac{1}{729}\)
=> \(3y-y=3+\frac{1}{1}+\frac{1}{3}+..+\frac{1}{729}-\frac{1}{1}-\frac{1}{3}-...-\frac{1}{2187}\)
<=> 2y = 3- 1/2187
=> y = \(\frac{3-\frac{1}{2187}}{2}\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)
=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)
Đặt A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
3A = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
3A - A = (\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)) - (\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\))
2A = 1 - \(\frac{1}{729}\) = \(\frac{728}{729}\)
A = \(\frac{728}{729}:2=\frac{364}{729}\)
\(G=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3G=3+1+\frac{1}{3}+...+\frac{1}{3^4}\)
\(3G-G=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)
\(2G=3-\frac{1}{3^5}\)
\(2G=3-\frac{1}{243}\)
\(2G=\frac{729}{243}-\frac{1}{243}\)
\(G=\frac{728}{243}:2\)
\(G=\frac{364}{243}\)
\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{x.\left(x+1\right)}=\frac{6042}{2015}\)
\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{6042}{2015}\)
\(1-\frac{1}{x+1}=\frac{6042}{2015}:3\)
\(1-\frac{1}{x-1}=\frac{2014}{2015}\)
\(\frac{1}{x-1}=1-\frac{2014}{2015}\)
\(\frac{1}{x-1}=\frac{1}{2015}\)
\(\Rightarrow x-1=2015\)
\(\Rightarrow x=2016\)
a, Gọi biểu thức đó là A
Ta có :
A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
A x 3 = \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}-\frac{1}{729}\)
A x 3 = \(1+A-\frac{1}{729}\)
A x 3 = \(\frac{728}{729}+A\)
A x 2 + A = \(\frac{728}{729}+A\)
A x 2 = \(\frac{728}{729}\)(bỏ A ở cả 2 vế)
A = \(\frac{728}{729}\div2=\frac{364}{729}\)
Đáp án = \(\frac{364}{729}\)
b, Phần này mình nghĩ là bạn sai đề rồi. Phải là \(\frac{45\times16-17}{45\times15+28}\)
1, \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{8}\right):\left(12\frac{1}{9}\right)=0\)
\(\left(\frac{6.9+5}{9}+x-\frac{117}{8}\right):\frac{12.9+1}{9}=\frac{1}{2}\)
( . là nhân nha)
\(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)
\(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)
\(x=\frac{113}{8}\)
( \(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)
\(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{49}{81}=\frac{56}{81}\)
\(4y=\frac{7}{81}\)
y = 7/81:4
y = 7/324