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Đặt \(D=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}+\frac{1}{2187}\)
\(\Leftrightarrow D=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}\)
\(\Leftrightarrow3D=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Leftrightarrow3D-D=2D=1-\frac{1}{3^6}\)
\(\Leftrightarrow D=\left(1-\frac{1}{3^6}\right)\div2\)
a=1 +1/3 +1/3^2 +1/3^3 +1/3^4 +1/3^5+1/3^6
3a=3 +1 +1/3 +1/3^2 + 1/3^3 +...+1/3^5
3a -a=[3 +1 +1/3 +1/3^2 +...+1/3^5] -1 -1/3 -1/3^2 -.........-1/3^6
2a =3- 1/3^6
a=[3-1/3^6] :2
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
nhân cả 2 vế với 3 ta có:
sx3=3+1+1/3 +1/9 +1/27 +1/81 +1/243
sx3-s=3 -1/729=2186/729
sx2=2186/729
s=2186/729 :2
s=1093/729
a) Cho: \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
\(\Rightarrow3A=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)
\(\Rightarrow3A-A=3-\frac{1}{81}\)
\(\Rightarrow A=\frac{3-\frac{1}{81}}{2}\)
\(A=\frac{121}{81}\)
b) \(37,52+4,7\times2,3-9,8\)
\(=37,52+10,81-9,8\)
\(=38,53\)
Chúc bn học tốt !!!!!
\(G=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(G=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3G=3+1+\frac{1}{3}+...+\frac{1}{3^4}\)
\(3G-G=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)
\(2G=3-\frac{1}{3^5}\)
\(2G=3-\frac{1}{243}\)
\(2G=\frac{729}{243}-\frac{1}{243}\)
\(G=\frac{728}{243}:2\)
\(G=\frac{364}{243}\)
\(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{x.\left(x+1\right)}=\frac{6042}{2015}\)
\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{6042}{2015}\)
\(1-\frac{1}{x+1}=\frac{6042}{2015}:3\)
\(1-\frac{1}{x-1}=\frac{2014}{2015}\)
\(\frac{1}{x-1}=1-\frac{2014}{2015}\)
\(\frac{1}{x-1}=\frac{1}{2015}\)
\(\Rightarrow x-1=2015\)
\(\Rightarrow x=2016\)
Đặt biểu thức trên là A ta có:
A = \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{12}\)+ \(\frac{1}{24}\)+ \(\frac{1}{48}\)+ \(\frac{1}{96}\)
A x 3 = \(1\)+ \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ \(\frac{1}{16}\)+ \(\frac{1}{32}\)
A x 3 = \(1\)+ \(1\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{8}\)+ \(\frac{1}{8}\)- \(\frac{1}{16}\)+ \(\frac{1}{16}\)- \(\frac{1}{32}\)
A x 3 = 2 - \(\frac{1}{32}\)= \(\frac{63}{32}\)
A = \(\frac{63}{32}\): 3 = \(\frac{63}{96}\)
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 = \(\frac{728}{729}:2=\frac{364}{729}\)
có cần phải giải ra ko
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(=\frac{81}{243}+\frac{27}{243}+\frac{9}{243}+\frac{3}{243}+\frac{1}{243}\)
\(=\frac{121}{243}\)
mk ko bít đúng hay ko nữa có gì mấy bạn góp ý cho mình nhé ! Thanks