91/243

91/243 đó em

Mình cần lời giải nhé

820/2187

3xB=3x(1/4+1/12+1/36+1/108+1/324+1/972+1/2916+1/8748)

3xB=3/4 + 1/4 +1/12 +1/36 +.........+1/2916

3xB - B= (3/4 + 1/4 + 1/12+1/36 + .........+1/2916) - ( 1/4 +1/12 +1/36 +1/108 + 1/324 + 1/972 + 1/2916 +1/8748 )

2xB =3/4 - 1/8748

2xB =1640/2187

B = 1640/2187 :2

B = 820/2187.

mình có cách làm như bn Kaito Kid và = 820/2187

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}.\)

\(3A=3\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\right).\)

\(3A=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+\frac{3}{108}+\frac{3}{324}+\frac{3}{972}\)

\(3A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

\(2A=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}-\frac{1}{4}-\frac{1}{12}-\frac{1}{36}-\frac{1}{108}-\frac{1}{324}-\frac{1}{972}\)

\(2A=\frac{3}{4}-\frac{1}{972}=\frac{182}{243}\)

\(\Rightarrow A=\frac{182}{243}:2=\frac{91}{243}\)

\(\Rightarrow A=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}=\frac{91}{243}\)

ta có    

A x 3 =3/4 + 3/12 + 3/36 +3/108 + 3/324 +3/972

A x 3=3/4 +1/4 + 1/12 +1/36 +1/108

A x 3 - A =(3/4 +1/12+1/36 +1/108)-(1/4 +1/12 +1/36 +1/108 +1/324 + 1/972)

A x 2=3/4 +1/12 +1/36 +1/108 - 1/4 -1/12 -1/36-1/108 -1/324 -1/972

A x 2= 3/4 - 1/972

A x 2= 728/972

A =728/972 : 2

A=91/243

\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)

=>\(2A=1+\frac{1}{2}+...+\frac{1}{2^8}\)

=>\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)\)

                       \(=1-\frac{1}{2^9}=\frac{511}{512}\)

\(B=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}+\frac{1}{972}\)

=>\(3B=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\frac{1}{324}\)

=>\(3B-B=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{324}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{972}\right)\)

=>\(2B=\frac{3}{4}-\frac{1}{972}=\frac{182}{243}\)

=>\(B=\frac{182}{243}:2=\frac{91}{243}\)

\(3C=1+\frac{1}{3}+.....+\frac{1}{3^{2015}}\)

\(\Rightarrow3C-C=2C=\left(1+\frac{1}{3}+.....+\frac{1}{3^{2014}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}.....+\frac{1}{3^{2015}}\right)=1-\frac{1}{3^{2015}}\)

\(\Rightarrow C=\frac{3^{2015}-1}{3^{2015}.2}\)

\(D=4\left(1+\frac{1}{3}+....+\frac{1}{3^5}\right)\)

\(\Rightarrow3D=4\left(3+1+....+\frac{1}{3^4}\right)\)

\(\Rightarrow3D-D=2D=4\left(3+1+....+\frac{1}{3^4}\right)-4\left(1+\frac{1}{3}+....+\frac{1}{3^5}\right)\)

\(\Rightarrow2D=4\left(3-\frac{1}{3^5}\right)\Rightarrow D=2\left(3-\frac{1}{3^5}\right)\)

A=511/512

B=91/243

A=511/512

B=91/243

A=511/512

B=91/243

a) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)

\(=5.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(=5.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right):2\)

\(=5.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)

\(=5.\left(1-\frac{1}{101}\right):2=5.\frac{100}{101}:2=\frac{500}{101}.\frac{1}{2}\)\(=\frac{250}{101}\)

b) \(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)

\(=3\left(\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\right)\)\(.\frac{1}{3}\)

\(=(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{30.33}).\frac{1}{3}\)

\(=(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}).\frac{1}{3}\)

\(=(\frac{1}{3}-\frac{1}{33}).\frac{1}{3}=\frac{10}{33}.\frac{1}{3}=\frac{10}{99}\)

câu c bạn có thể viết rõ được ko

Làm nốt câu c hộ các bạn:

\(\frac{\frac{3}{41}-\frac{12}{47}+\frac{27}{53}}{\frac{4}{41}-\frac{16}{47}+\frac{36}{53}}=\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)

\(a)\) \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

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

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)

\(A=1-\frac{1}{2^9}\)

\(A=\frac{2^9-1}{2^9}\)

Vậy \(A=\frac{2^9-1}{2^9}\)

Chúc bạn học tốt ~ 

Đặt \(S=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{2010}+\frac{1}{6030}.\)

        \(\Rightarrow3S=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+...+\frac{3}{2010}+\frac{3}{6030}\)

                   \(=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{670}+\frac{1}{2010}\)

  \(\Rightarrow3S-S=2S=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{670}+\frac{1}{2010}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{2010}+\frac{1}{6030}\right)\)

\(2S=\frac{3}{4}-\frac{1}{6030}\)

\(\Rightarrow S=\frac{\frac{3}{4}-\frac{1}{6030}}{2}\)