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NN
2
10 tháng 5 2017
\(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\)
\(=3\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\)
Đặt \(A=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)
\(\Rightarrow2A=2+1+...+\dfrac{1}{2^8}\)
\(\Rightarrow2A-A=\left(2+1+...+\dfrac{1}{2^8}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\right)\)
\(\Rightarrow A=2-\dfrac{1}{2^9}\)
\(\Rightarrow S=3\left(2-\dfrac{1}{2^9}\right)=6-\dfrac{3}{2^9}\)
Vậy...
10 tháng 5 2017
=> S = 3 \(\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^9}\right)\)
S = 3 \(\left(2+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{2^2}+...+\dfrac{1}{2^8}-\dfrac{1}{2^9}\right)\)
= 3\(\left(2-\dfrac{1}{2^9}\right)\)= 6 \(-\dfrac{3}{2^9}\)
TICK CHO MK NHA!
NT
2
PT
28 tháng 3 2017
Giải:
\(S=3.\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\)
\(2S=3.\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)\)
\(2S-S=3.\left[\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^8}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\right]\)
\(S=3.\left(2-\dfrac{1}{2^9}\right)\)
\(S=3.\dfrac{1023}{512}\)
\(S=\dfrac{3069}{512}\)
28 tháng 3 2017
2.S= \(6\) + \(3\) + \(\dfrac{3}{2}\) + ..... + \(\dfrac{3}{2^8}\)
2.S - S= ( \(6\) + \(3\) + \(\dfrac{3}{2}\) + ..... + \(\dfrac{3}{2^8}\)) - (\(3\) + \(\dfrac{3}{2}\) + \(\dfrac{3}{2^2}\) + ..... + \(\dfrac{3}{2^9}\))
S= 6 - \(\dfrac{3}{2^9}\)
S= \(\dfrac{6.512}{512}\) - \(\dfrac{3}{512}\) = \(\dfrac{3069}{512}\)
Bn tự rút gọn nha, mk hơi nhát
NT
3
DT
Đỗ Thanh Hải
CTVVIP
8 tháng 5 2017
Ta có:
S = 3 + \(\dfrac{3}{2}\) + \(\dfrac{3}{2^2}\) + ........+ \(\dfrac{3}{2^9}\)
2S = 6 + 3 + \(\dfrac{3}{2^2}\) + ....... + \(\dfrac{3}{2^8}\)
Mà S = 3 + \(\dfrac{3}{2}\) + \(\dfrac{3}{2^2}\) + ........+ \(\dfrac{3}{2^9}\)
=> 2S - S = 6 - \(\dfrac{3}{2^9}\)
S = \(\dfrac{3069}{512}\)
NT
7 tháng 5 2017
\(S=3\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\right)\)
Đặt : \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\\ \Rightarrow2.A=2+1+....+\dfrac{1}{2^8}\\ \Rightarrow A=2-\dfrac{1}{2^9}\\ \Rightarrow S=3\left(2-\dfrac{1}{2^9}\right)\)
NT
1
2 tháng 5 2017
Ta có :
\(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+..............+\dfrac{3}{2^9}\)
\(\Rightarrow2S=2\left(3+\dfrac{3}{2}+\dfrac{3}{2^2}+..............+\dfrac{3}{2^9}\right)\)
\(\Rightarrow2S=6+3+\dfrac{3}{2}+\dfrac{3}{2^2}+.............+\dfrac{3}{2^8}\)
\(\Rightarrow2S-S=\left(6+3+\dfrac{3}{2}+\dfrac{3}{2^2}+.......+\dfrac{3}{2^8}\right)-\left(3+\dfrac{3}{2}+\dfrac{3}{2^2}+......+\dfrac{3}{2^9}\right)\)
\(\Rightarrow S=6-\dfrac{3}{2^9}\)
\(\Rightarrow S=6-\dfrac{3}{512}=\dfrac{3069}{512}\)
Q
17 tháng 4 2017
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
13 tháng 4 2017
\(S>\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{9.10}\)
\(S>\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\) (1)
\(S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{8.9}\)
\(S< 1-\dfrac{1}{9}=\dfrac{8}{9}\) (2)
(1) và (2) => đpcm
8 tháng 4 2017
Ta có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{2^2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)\(=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{9}=\dfrac{23}{36}< \dfrac{32}{36}=\dfrac{8}{9}\). (1)
Ta lại có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{19}{20}>\dfrac{8}{20}=\dfrac{2}{5}\). (2)
Từ (1) và (2) suy ra đpcm.
26 tháng 3 2017
b, B = 1 + 2 + 2^2 + 2^3 +.....+ 2^2013
2B = 2.(1 + 2 + 2^2 + 2^3 +.....+ 2^2013)
2B = 2 + 2^2 + 2^3 + 2^4 +.....+ 2^2014
2B - B = 2^2014 - 1
B = 2^2014 - 1
8 tháng 5 2017
Câu a :
Chưa nghĩ ra! Sorry nhé!!
Câu b :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Câu c :
Câu hỏi của Trần Thùy Linh - Toán lớp 6 | Học trực tuyến
Vào link đó mà xem, t ngại chép lại
Ta có:
\(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\)
\(\Rightarrow2S=2\left(3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\right)\)
\(\Rightarrow2S=6+3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^8}\)
\(\Rightarrow2S-S=\left(6+3+\dfrac{3}{2}+...+\dfrac{3}{2^8}\right)-\left(3+\dfrac{3}{2}+\dfrac{3}{2^2}+...+\dfrac{3}{2^9}\right)\)
\(\Rightarrow S=6-\dfrac{3}{2^9}=6-\dfrac{3}{512}=\dfrac{3069}{512}\)
Vậy \(S=\dfrac{3069}{512}\)