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DA
16 tháng 8 2017
a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)
\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)
\(\frac{1}{4}xA=\frac{127}{384}\)
\(A=\frac{127}{384}:\frac{1}{4}\)
\(A=\frac{127}{96}\)
KT
15 tháng 10 2018
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
NS
1
29 tháng 10 2018
mình xin nhầm đề là: A = \(1*2*3+2*4*6+4*8*12+8*16*24 \over2*3*4+4*6*8+8*12*16+16*24*32\)
LB
11 tháng 5 2017
a)
\(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(\frac{2.3\left(1.2\right)+2.3\left(2.4\right)+2.3\left(3.6\right)+2.3\left(4.8\right)+2.3\left(5.10\right)}{3.4\left(3.4+6.8+9.12+12.16+15.20\right)}\)
\(=\frac{\left(3.4+6.8+9.12+12.16+15.20\right)}{2.3\left(3.4+6.8+9.12+12.16+15.20\right)}=\frac{1}{2.3}=\frac{1}{6}\)
XO
23 tháng 7 2019
1 a) (x+ 1) + (x + 2 ) + (x + 3) + ... + (x + 100) = 205550 (100 cặp)
=> (x + x + ... + x) + (1 + 2 + 3 + ... + 100) = 205 550
100 số hạng x 100 số hạng
=> 100.x + 100 . 101 : 2 = 205 550
=> 100.x + 5050 = 205 550
=> 100 . x = 205 550 - 5050
=> 100 . x = 200500
=> x = 200500 : 100
=> x = 2005
NT
4
18 tháng 7 2015
bạn cứ đọc kkĩ lại đi, tại sao chỗ D=2*4*+ lại có cả dấu + lẫn dấu nhân
7 tháng 8 2016
2*4+2*4*8+4*8*16+8*16*32/3*4+2*6*8+4*12*16+8*24*32
bai toan tinh nhanh
VN
4
29 tháng 4 2019
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}+\frac{1}{768}+\frac{1}{1536}\)
\(A\times2=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}+\frac{2}{384}+\frac{2}{768}+\frac{2}{1536}\)
Rút gọn ta được
\(A\times2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}+\frac{1}{768}\)
\(A\times2-A=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{768}-\left[\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1536}\right]\)
\(A=\frac{2}{3}+\frac{1}{3}-\frac{1}{3}-\frac{1}{1536}\)
\(A=\frac{2}{3}-\frac{1}{1536}=\frac{341}{512}\)
29 tháng 3 2023
Ta có:
\(S=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}+\dfrac{2}{384}\\ =\dfrac{2}{3}+\dfrac{2}{2\times3}+\dfrac{2}{2\times6}+\dfrac{2}{2\times12}+\dfrac{2}{2\times24}+\dfrac{2}{2\times48}+\dfrac{2}{2\times96}+\dfrac{2}{2\times192}\\ =\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}\\ \)
\(\dfrac{S}{2}=\dfrac{1}{2}\left(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}+\dfrac{2}{384}\right)\\ =\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}+\dfrac{1}{384}\)
\(S-\dfrac{S}{2}=\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}-\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}+\dfrac{1}{384}\right)\\ =\dfrac{2}{3}-\dfrac{1}{384}=\dfrac{2\times128-1}{384}\\ =\dfrac{85}{128}\\ \Rightarrow S=\dfrac{85}{128}\times2=\dfrac{85}{64}\)
29 tháng 3 2023
\(A=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}+\dfrac{2}{384}\)
\(A.2=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\)
\(A=A.2-A=\dfrac{4}{3}-\dfrac{2}{384}=\dfrac{127}{96}\)