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DD
Đoàn Đức Hà
Giáo viên
2 tháng 6 2021
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}+\frac{10-9}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)=7,6\)
b) Bạn làm tương tự.
KT
15 tháng 10 2018
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}+1-\frac{1}{110}\)
\(=10-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\right)\)
\(=10-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{10}\right)\)
\(=\frac{91}{10}\)
K
1
28 tháng 3 2015
1/2 + 5/6 + 11/12 + ... + 89/90 + 109/110 + 10/11
= (1 - 1/2) + (1 - 1/6) + (1 - 1/12) + ... + (1 - 1/90) + (1 - 1/110) + 10/11
= (1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) - (1/1*2 + 1/2*3 + 1/3*4 + ... + 1/10*11) + 10/11
= 9 - (1 - 1/11) + 10/11
= 9 - 10/11 + 10/11 = 9
PH
2
3 tháng 12 2019
Bài làm
a) 11 - 12 + 13 - 14 + 15 - 16 + 17 - 18 + 19 - 20
= ( 11 - 12 ) + ( 13 - 14 ) + ( 15 - 16 ) + ( 17 - 18 ) + ( 19 - 20 )
= ( -1 ) + ( -1 ) + ( -1 ) + ( -1 ) + ( -1 )
= -5
b) 101 - 102 - ( - 103 ) - 104 - ( - 105 ) - 106 - ( - 107 ) - 108 - ( - 109 ) - 110
= 101 - 102 + 103 - 104 + 105 - 106 + 107 - 108 + 109 - 110
= ( 101 - 102 ) + ( 103 - 104 ) + ( 105 - 106 ) + ( 107 - 108 ) + ( 109 - 110 )
= ( -1 ) + ( -1 ) + ( -1 ) + ( -1 ) + ( -1 )
= -5
# Học tốt #
KD
3 tháng 12 2019
a)11 - 12 + 13 - 14 + 15 - 16 + 17 - 18 + 19 - 20
= (11 - 12) + (13 - 14) + (15 - 16) + (17 - 18) + ( 19 - 20)
= (-1) + (-1) + (-1) + (-1) + (-1) = (-1) . 5 = - 5
b)101-102-(-103)-104-(-105)-106-(-107)-108-(-109)-110
= 101-102+103-104+105-106+107-108+109-110
= (101-102)+(103-104)+(105-106)+(107-108)+(109-110)
= (-1) + (-1) + (-1) + (-1) + (-1)
= (-1) . 5
= -5
LH
1
14 tháng 7 2015
A - B = \(\left(1+\frac{1}{2}+1+\frac{1}{12}+1+\frac{1}{30}+1+\frac{1}{56}+1+\frac{1}{90}\right)-\left(1-\frac{1}{6}+1-\frac{1}{20}+1-\frac{1}{42}+1-\frac{1}{72}+1-\frac{1}{110}\right)\)= \(\left(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)-\left(5-\frac{1}{6}-\frac{1}{20}-\frac{1}{42}-\frac{1}{72}-\frac{1}{110}\right)\)\
= \(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}-5+\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\)
= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\)
29 tháng 12 2015
a,11-12+13-14+15-16+17-18+19-20
=(11-12)+(13-14)+(15-16)+(17-18)+(19-20)
=(-1)+(-1)+(-1)+(-1)+(-1)
=-5
b,101-102-(-103)-104-(-105)-106-(-107)-108-(-109)-110
=101-102+103-104+105-106+107-108+109-110
=(101-102)+(103-104)+(105-106)+(107-108)+(109-110)
=(-1)+(-1)+(-1)+(-1)+(-1)
=-5
Đặt biểu thức là A. A có 10 số hạng.
A = 1/2+5/6+11/12+19/20+...+109/110.
A = (1-1/2) + (1-1/6) + ...+(1-1/110)
A = 1+1+1+...+1(10 số 1) - (\(\frac{1}{2}\)+\(\frac{1}{6}\)+...+\(\frac{1}{110}\))
A=10-B
B = \(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{10.11}\)
B = \(\frac{2-1}{1.2}\)+\(\frac{3-2}{2.3}\)+\(\frac{4-3}{3.4}\)+...+\(\frac{11-10}{10.11}\)
B=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{10}\)-\(\frac{1}{11}\)
B=1-\(\frac{1}{11}\)=\(\frac{10}{11}\)
⇒A=10-B=10-\(\frac{10}{11}\)=\(\frac{100}{11}\)