Cho \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)

    \(\frac{1}{3}A=\frac{1}{3}\times\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\right)\)

    \(\frac{1}{3}A=\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{19683}\)

 \(A-\frac{1}{3}A=\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{6561}\right)-\left(\frac{1}{9}+\frac{1}{27}+...+\frac{1}{19683}\right)\)

\(\frac{2}{3}A=\frac{1}{3}-\frac{1}{19683}\)

\(A=\frac{4840}{9683}:\frac{2}{3}=\frac{7260}{9683}\)

a)

\(\begin{array}{l}x.\frac{{14}}{{27}} = \frac{{ - 7}}{9}\\x = \frac{{ - 7}}{9}:\frac{{14}}{{27}}\\x = \frac{{ - 7}}{9}.\frac{{27}}{{14}}\\x = \frac{{ - 3}}{2}\end{array}\)                

Vậy \(x = \frac{{ - 3}}{2}\).

b)

\(\begin{array}{l}\left( {\frac{{ - 5}}{9}} \right):x = \frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right):\frac{2}{3}\\x = \left( {\frac{{ - 5}}{9}} \right).\frac{3}{2}\\x = \frac{{ - 5}}{6}\end{array}\)

Vậy \(x = \frac{{ - 5}}{6}\).

c)

\(\begin{array}{l}\frac{2}{5}:x = \frac{1}{{16}}:0,125\\\frac{2}{5}:x = \frac{1}{{16}}:\frac{1}{8}\\\frac{2}{5}:x = \frac{1}{{16}}.8\\\frac{2}{5}:x = \frac{1}{2}\\x = \frac{2}{5}:\frac{1}{2}\\x = \frac{2}{5}.2\\x = \frac{4}{5}\end{array}\)      

Vậy \(x = \frac{4}{5}\)

d)

\(\begin{array}{l} - \frac{5}{{12}}x = \frac{2}{3} - \frac{1}{2}\\ - \frac{5}{{12}}x = \frac{4}{6} - \frac{3}{6}\\ - \frac{5}{{12}}x = \frac{1}{6}\\x = \frac{1}{6}:\left( { - \frac{5}{{12}}} \right)\\x = \frac{1}{6}.\frac{{ - 12}}{5}\\x = \frac{{ - 2}}{5}\end{array}\)

Vậy \(x = \frac{{ - 2}}{5}\).

Chú ý: Khi trình bày lời giải bài tìm x, sau khi tính xong, ta phải kết luận.

\(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}\)

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)

\(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)

\(2B=1-\frac{1}{3^8}\)

\(B=\frac{1-\frac{1}{3^8}}{2}\)

B = 1/3 + 1/9 + 1/27 + ... + 1/6561 

B = 1/3^1 + 1/3^2 + 1/3^3 + ... + 1/3^8 

3B = 1 + 1/3^1 + 1/3^2 + ... + 1/3^7 

3B - B = ( 1 + 1/3^1 +1/3^2 + ... + 1/3^7 ) - ( 1/3^1 + 1/3^2 + 1/3^3 + .... + 1/3^8 ) 

2B = 1 - 1/3^8

B = 1 - 1/3^8  / 2

làm theo cách lớp 6 hộ mik vs

\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+...+\left(a+\frac{1}{23.25}\right)=11a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)

\(\Rightarrow12a+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11a+\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)\)(1)

Ta có \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)=\frac{1}{2}\left(1-\frac{1}{25}\right)=\frac{1}{2}.\frac{24}{25}=\frac{12}{25}\)

Lại có \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}=\frac{3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)}{2}\)

\(=\frac{1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-\frac{1}{3^4}-\frac{1}{3^5}}{2}=\frac{1-\frac{1}{3^5}}{2}=\frac{1}{2}-\frac{1}{3^5.2}\)

Khi đó (1) <=> \(12a-\frac{12}{25}=11a+\frac{1}{2}-\frac{1}{3^5.2}\)

=> \(a=\frac{12}{25}+\frac{1}{2}-\frac{1}{3^5.2}=\frac{49}{50}-\frac{1}{3^5.2}=\frac{49}{50}-\frac{1}{486}=\frac{23764}{24300}\)

Gọi \(A=\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)\)

\(\Rightarrow A=12a+\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{23.25}\right)\)

\(\Rightarrow A=12a+\left[\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{23.25}\right)\right]\)

\(\Rightarrow A=12a+\left[\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]\)

\(\Rightarrow A=12a+\left[\frac{1}{2}\left(1-\frac{1}{25}\right)\right]\)

\(\Rightarrow A=12a+\left(\frac{1}{2}.\frac{24}{25}\right)\)

\(\Rightarrow A=12a+\frac{12}{25}\)

Gọi \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)

\(\Rightarrow B=\frac{1}{1.3}+\frac{1}{3.3}+\frac{1}{9.3}+\frac{1}{27.3}+\frac{1}{81.3}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

\(\Rightarrow3B-B=1-\frac{1}{243}\)

\(\Rightarrow2B=\frac{242}{243}\)

\(\Rightarrow B=\frac{121}{243}\)

\(\Rightarrow A=11a+B\)

\(\Rightarrow12a+\frac{12}{25}=11a+\frac{121}{243}\)

\(\Leftrightarrow12a-11a=\frac{121}{243}-\frac{12}{25}\)

\(\Leftrightarrow a=\frac{109}{6075}\)

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

Ta thấy tất cả phân số này đều có mẫu chung là 81

=> A = \(\frac{81}{81}\)+ \(\frac{27}{81}\)+ \(\frac{9}{81}\)+\(\frac{3}{81}\)+ \(\frac{1}{81}\)( lấy 81 chia cho mẫu rồi nhân cho tử, đặt mẫu số là 81 )

=> A = \(\frac{81+27+9+3+1}{81}\)= \(\frac{121}{81}\)

nhớ ủng hộ mik với nha mn

=121/81 nhé bạn

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

\(=\frac{81}{81}+\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\)

  \(=\frac{121}{81}\)

\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{531441}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{12}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{11}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{11}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{12}}\right)\)

\(2A=1-\frac{1}{3^{12}}\)

\(2A=\frac{531440}{531441}\)

\(A=\frac{531440}{531441}\div2\)

\(A=\frac{265720}{531441}\)

Chúc bạn học tốt!!!!!!!!

a)

\(\Rightarrow A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)

\(\Rightarrow A=\frac{1}{5}+\frac{2}{7}\)

\(\Rightarrow A=\frac{17}{35}\)

b)

\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{56}-\frac{1}{61}\right)\)

\(\Rightarrow B=5\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(\Rightarrow B=5.\frac{50}{671}=\frac{250}{671}\)

c)

\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)

\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)

\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)

\(\Rightarrow C=1-1-\frac{1}{25}\)

\(\Rightarrow C=\frac{1}{25}\)

A>5/3>5/4=>A>5/4 chứ mị

mk nhìn nhầm