Tính hợp lí, nếu có thể
a) 45.18+18.53 b) ( 7/8+1/2):11/16 c) 1/6+1/12+1/20+1/30+......+1/90+1/110
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KV
20 tháng 3 2023
a) 2/9 - (1/20 + 2/9)
= 2/9 - 1/20 - 2/9
= (2/9 - 2/9) - 1/20
= 0 - 1/20
= -1/20
b) -3/14 + 2/13 + (-25/14) + (-15/13)
= (-3/14 - 25/14) + (2/13 - 15/13)
= -2 - 1
= -3
c) -3/11 + 11/8 - 3/8 + (-8/11)
= (-3/11 - 8/11) + (11/8 - 3/8)
= -1 + 1
= 0
d) 3/8 + (-1/4) - (7/12 - 1/6)
= 1/8 - 5/12
= -7/24
e) (1/3 + 12/67 + 13/41) - (79/67 - 28/41)
= 1/3 + 12/67 + 13/41 - 79/67 + 28/41
= 1/3 + (12/67 - 79/67) + (13/41 + 28/41)
= 1/3 - 1 + 1
= 1/3
S
1
19 tháng 6 2015
\(C=\frac{8}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{8}{90}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
\(=\frac{8}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(=\frac{4}{45}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{4}{45}-\left(1-\frac{1}{9}\right)=\frac{4}{45}-\frac{8}{9}=\frac{4}{45}-\frac{40}{45}=\frac{-36}{45}=\frac{-4}{5}\)
B
2
4 tháng 2 2016
a.A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
b. 1/2 + 1/6 + 1/12 + … + 1/110
= 1/1.2 + 1/2.3 + 1/3.4 + … + 1/10.11. (dấu . thay dấu x).
= 1/1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…+ 1/10 – 1/11
= 1/1 – 1/11
= 10/11
Chúc bạn học giỏi nha!
4 tháng 2 2016
a ) Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}\)
\(\Rightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
Nhân 2 vào hai vế của biểu thức A , ta được :
\(2A=2.\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(\Rightarrow2A=1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
Lấy biểu thức 2A - A , ta được :
\(2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+....+\frac{1}{2^8}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(\Rightarrow A=1-\frac{1}{2^9}\Rightarrow A=\frac{512}{512}-\frac{1}{512}=\frac{511}{512}\)
Vậy \(A=\frac{511}{512}\)
b ) Đặt \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}+\frac{1}{110}\)
\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Rightarrow B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(B=1-\frac{1}{11}=\frac{11}{11}-\frac{1}{11}=\frac{10}{11}\)
Vậy \(B=\frac{10}{11}\)
29 tháng 8 2023
a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)
=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)
=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)
=>\(A=\dfrac{127}{128}\)
b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
18 tháng 9 2023
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)
17 tháng 9 2023
\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)
\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)
\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)
NT
7
a, có người giải ùi khỏi làm nữa
b, (7/8 + 1/2) : 11/16
= 7/8 : 11/16 + 1/2 : 11/16
= 14/11 + 8/11
= 2
c, \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)
= \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{10}-\frac{1}{11}\)
= \(\frac{1}{2}-\frac{1}{11}\)
= \(\frac{9}{22}\)
2NHEBAN