\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+....+\frac{1}{59049}\)
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tính nhanh
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+....\frac{1}{59049}\)
Đặt \(A=\frac{1}{3}+\frac{1}{9}+.......+\frac{1}{59049}\)
\(3A=3.\left(\frac{1}{3}+\frac{1}{9}+......+\frac{1}{59049}\right)\)
\(3A=1+\frac{1}{3}+........+\frac{1}{19683}\)
\(3A-A=\left(1+\frac{1}{3}+......+\frac{1}{19683}\right)-\left(\frac{1}{3}+\frac{1}{9}+........+\frac{1}{59049}\right)\)
\(2A=1-\frac{1}{59049}\)
\(2A=\frac{59048}{59049}\)
\(A=\frac{59048}{59049}:2\)
\(A=\frac{59048}{118098}\)
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\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+......+\frac{1}{59049}\)
Tính biểu thức trên
A=$\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+......+\frac{1}{59049}$
3A=$\frac{1}+frac{1}{3}+\frac{1}{9}+\frac{1}{27}+......+\frac{1}{19683}$
3A-A=2A=1-1/59049=59048/59049
A=59048/118098
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Tính nhanh:
8-\(\frac{1}{3}\)+\(\frac{1}{9}\)+\(\frac{1}{27}\)+\(\frac{1}{81}\)+............+\(\frac{1}{59049}\)
\(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{-3}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)
\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
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
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\(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81} \)
\(=\frac{81}{81}+\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\)
\(=\frac{121}{81}\)
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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)
\(\Rightarrow C=1-\left(\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+....+\frac{1}{49.25}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{49.50}\right)\)
\(\Rightarrow C=1-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(\Rightarrow C=1-1-\frac{1}{25}\)
\(\Rightarrow C=\frac{1}{25}\)
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Tính nhanh:
\(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\text{Đặt : }A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(\Rightarrow3A-A=1-\frac{1}{729}\)
\(\Rightarrow2A=\frac{728}{729}\)
\(\Rightarrow A=\frac{728}{729}:2=\frac{364}{729}\)
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\(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\)
\(\Rightarrow2A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}\)
\(\Rightarrow2A-A=\frac{1}{3^1}-\frac{1}{3^7}\)
\(\Rightarrow A=\frac{1}{3^1}-\frac{1}{3^7}\)
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\(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+\frac{1}{729}\)
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+...+\frac{1}{531441}\)
\(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+...+\frac{1}{531441}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{12}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{11}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{11}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{12}}\right)\)
\(2A=1-\frac{1}{3^{12}}\)
\(2A=\frac{531440}{531441}\)
\(A=\frac{531440}{531441}\div2\)
\(A=\frac{265720}{531441}\)
Chúc bạn học tốt!!!!!!!!
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