Tính tổng:S=5/6+11/12+19/20+...+89/90
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5/6 + 11/12 + 19/20 + ... + 89/90
= 1 - 1/6 + 1 - 1/12 + 1 - 1/20 + ... + 1-1/90
= [1+1+1+1...+1] - [1/2*3 + 1/3*4 + 1/4*5 + ... +1/9*10]
= 8 - [1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/9 - 1/10]
= 8 - [1/2 - 1/10]
= 8 - 2/5
= 38/5
S=1-1/6+1-1/12+...+1-1/90
=8-(1/2.3+1/3.4+...+1/9.10)
=8-(1/2-1/3+1/3-1/4+...+1/9-1/10)
=8-(1/2-1/10)
=8-2/5
=38/5
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+.....+\frac{71}{72}+\frac{89}{90}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+....+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(=\left(1+1+1+....+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}=\frac{81}{10}\)
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{90}\right)\\ =\left(1+1+1+1+1+1+1+1+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\\ =9-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\\ =9-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =9-\left(1-\dfrac{1}{10}\right)=9-\dfrac{9}{10}=\dfrac{81}{10}\)
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=8-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=8-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{38}{5}\)
Đặt A = 1/2 + 5/6 + 11/12 + 19/20 + ... + 89/90
A = ( 1 - 1/2 ) + ( 1 - 1/12 ) + ( 1 - 1/20 ) + ... + ( 1 - 1/90 )
A = ( 1 + 1 + 1 + ... + 1 + 1 ) - ( 1/2 + 1/6 + 1/20 + ... + 1/90
A = 9 - ( 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10 )
A = 9 - ( 1 - 1/10 )
A = 9 - 9/10
A = 81/10
Bạn ra câu đố mik xin trả lời như sau :
Ta có : A = \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
A có số các phân số là :
( 89 - 5 ) : 1 + 1 = 85 ( p/số )
Ta có : A = \(\left(1-\frac{5}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+...+\left(1-\frac{1}{90}\right)\)
A = \(\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
A = \(\left(1.85\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)( Có 85 số 1 vì mỗi số 1 đi kèm với 1 phân số mà có 85 phân số)
A = \(85-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
A = \(85-\left(\frac{1}{2}-\frac{1}{10}\right)\)
A = \(85-\frac{2}{5}\)
A = 85 - 0.4
A = 84,6
Ủng hộ mik nhé ^_^"
A = (1 -1/2) + (1 - 1/6) + (1 - 1/12) + (1 - 1/20 ) + ...+ (1 - 1/ 90)
= (1+1+1+1+1+1+1+1+1) - ( 1/2 - 1/6 - 1/12 - 1/ 20 - ...- 1/90)\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)\(=9-\left(1-\frac{1}{10}\right)=\frac{81}{10}\)
\(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+....+\frac{89}{90}\)
\(S=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+....+\left(1-\frac{1}{90}\right)\)
\(S=\left(1-\frac{1}{2.3}\right)+\left(1-\frac{1}{3.4}\right)+\left(1-\frac{1}{4.5}\right)+....+\left(1-\frac{1}{9.10}\right)\)
\(S=\left(1+1+1+....+1\right)-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(S=8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(S=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(S=8-\frac{2}{5}=\frac{38}{5}\)