\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
HK
Những câu hỏi liên quan
\(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{-3}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+...+\frac{1}{531441}\)
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{531441}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{12}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{11}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{11}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{12}}\right)\)
\(2A=1-\frac{1}{3^{12}}\)
\(2A=\frac{531440}{531441}\)
\(A=\frac{531440}{531441}\div2\)
\(A=\frac{265720}{531441}\)
Chúc bạn học tốt!!!!!!!!
Đúng 0
Bình luận (0)
A = \(\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
B = \(\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.81}\)
C = \(1-\frac{1}{3}-\frac{1}{6}-\frac{1}{10}-\frac{1}{15}-...-\frac{1}{1225}\)
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)
\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow C=1-1-\frac{1}{25}\)
\(\Rightarrow C=\frac{1}{25}\)
Đúng 0
Bình luận (0)
Tìm A,biết:
A = \(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+.........+\frac{1}{6561}\)
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{6561}\)
\(\Rightarrow A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^8}\)
\(\Rightarrow3A=3.\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\) \(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^8}\)
\(\Rightarrow2A=1-\frac{1}{3^8}\) \(\Rightarrow A=\frac{1-\frac{1}{3^8}}{2}\)
k cho mik đi mn!Nguyễn Như Quỳnh!
Đúng 0
Bình luận (0)
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+....+\frac{1}{59049}\)
Chứng minh rằng:frac{1}{5}+frac{1}{15}+frac{1}{25}+....+frac{1}{1985} frac{9}{20}mk làm thế này đúng ko mọi ngườiĐặt Afrac{1}{3}+frac{1}{5}+frac{1}{7}+frac{1}{9}+......+frac{1}{243}Afrac{1}{3}+left(frac{1}{5}+frac{1}{7}+frac{1}{9}right)+left(frac{1}{11}+frac{1}{13}+frac{1}{15}+....+frac{1}{27}right)+left(frac{1}{29}+frac{1}{31}+frac{1}{33}+....+frac{1}{81}right)+left(frac{1}{83}+frac{1}{85}+frac{1}{87}+.....+frac{1}{243}right)Afrac{1}{3}+frac{1}{9}.3+frac{1}{27}.9+frac{1}{81}.27+frac{1}{243}.81f...
Đọc tiếp
Chứng minh rằng:\(\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+....+\frac{1}{1985}< \frac{9}{20}\)
mk làm thế này đúng ko mọi người
Đặt \(A=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+......+\frac{1}{243}\)
\(A=\frac{1}{3}+\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}\right)+\left(\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+....+\frac{1}{27}\right)+\left(\frac{1}{29}+\frac{1}{31}+\frac{1}{33}+....+\frac{1}{81}\right)+\left(\frac{1}{83}+\frac{1}{85}+\frac{1}{87}+.....+\frac{1}{243}\right)\)
\(=>A>\frac{1}{3}+\frac{1}{9}.3+\frac{1}{27}.9+\frac{1}{81}.27+\frac{1}{243}.81=\frac{1}{3.5}=\frac{5}{3}\)
\(=>A>\frac{5}{3}>\frac{5}{4}=>A< \frac{5}{4}\)
\(=>\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+....+\frac{1}{397}< \frac{5}{4}\)
\(=>1+\frac{1}{3}+\frac{1}{7}+....+\frac{1}{397}< \frac{5}{4}\)
\(=>\frac{1}{5}.\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+....+\frac{1}{397}\right)< \frac{9}{4}.\frac{1}{5}\)
\(=>\frac{1}{5}+\frac{1}{15}+\frac{1}{25}+......+\frac{1}{1985}< \frac{9}{20}\)
cách tính nhanh phân số \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{19683}\)
Tính nhanh:
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\text{Đặt : }A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=1-\frac{1}{729}\)
\(\Rightarrow2A=\frac{728}{729}\)
\(\Rightarrow A=\frac{728}{729}:2=\frac{364}{729}\)
Đúng 0
Bình luận (0)
\(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow2A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}\)
\(\Rightarrow2A-A=\frac{1}{3^1}-\frac{1}{3^7}\)
\(\Rightarrow A=\frac{1}{3^1}-\frac{1}{3^7}\)
Đúng 0
Bình luận (0)
\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)