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2 tháng 1 2016

a, Dat A =\(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\)

\(\Rightarrow\frac{1}{3}A=\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\)

\(\Rightarrow\frac{1}{3}A+A=\left(\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-...-\frac{1}{3^{199}}+\frac{1}{3^{200}}\right)+\left(\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-...-\frac{1}{3^{198}}+\frac{1}{3^{199}}\right)\)

\(\Rightarrow\frac{4}{3}A=\frac{1}{3}+\frac{1}{3^{200}}\)

\(\Rightarrow A=\frac{\frac{1}{3}+\frac{1}{3^{200}}}{\frac{4}{3}}\)

chung minh tuong tu cau b va c

10 tháng 10 2020

1) Đặt \(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)

\(\Rightarrow3D=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3D-D=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)

\(\Leftrightarrow2D=1-\frac{1}{3^{100}}\)

\(\Leftrightarrow D=\frac{3^{100}-1}{2\cdot3^{100}}\)

Vậy \(D=\frac{3^{100}-1}{2\cdot3^{100}}\)

2) Ta có: \(\frac{49}{58}\cdot\frac{2^5}{4^2}-\frac{7^2}{-58}\cdot3\)

\(=\frac{49}{58}\cdot2-\frac{49}{58}\cdot3\)

\(=-1\cdot\frac{49}{58}\)

\(=-\frac{49}{58}\)

HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023

a) Cách 1:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)

Cách 2:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)

b)

\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)

25 tháng 8 2019

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\left[\frac{1}{199}+1\right]+\left[\frac{2}{198}+1\right]+\left[\frac{3}{197}+1\right]+...+\left[\frac{198}{2}+1\right]}\)

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}}\)

\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{200\left[\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right]}=\frac{1}{200}\)

30 tháng 3 2020

1)  A = -1 / 3

2)  B = -35 / 2

  Chúc em làm bài tốt !

15 tháng 2 2016

Để ý ta thấy 63.1,2-21.3,6+1=0

=>A=0

23 tháng 1 2016

đề ý 63.1,2-21.3,6+1=0=>B=0