Ket qua cua tong: A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 = ............
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a) 1 = 1
10 = 2 . 5
ƯCLN(1, 10) = 1
b) 11 = 11
15 = 3 . 5
ƯCLN(11, 15) = 1
c) 18 = 2 . 32
42 = 2 . 3 . 7
ƯCLN(18, 42) = 2 . 3 = 6
d) 24 = 23 . 3
16 = 24
ƯCLN(24, 16) = 23 = 8
a)ƯCLN (1,10)={1}
b)ƯCLN (11,15)={1}
c)Ta có :18=2.32
42=2.3.7
\(\Rightarrow\)ƯCLN(18,42)=2.3=6
d)Ta thấy: 24 ;16;8 \(⋮\)8 \(\Rightarrow\)ƯCLN (24;16;8)=8
A=1/2+1/6+....+1/56+1/72
A=1/1.2+1/2.3+...+1/7.8+1/8.9
A=1/1-1/2+1/2-1/3+...+1/7-1/8+1/8-1/9
A=1/1-1/9=9/9-1/9=8/9
A = \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}\)
A = \(\frac{1}{2}+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)
A = \(\left(1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
A = \(9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
A = \(9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
A = \(9-\left(1-\frac{1}{10}\right)=9-\frac{9}{10}\)
A = \(\frac{81}{10}\)
\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{89}{90}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)=10-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(=10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)=10-\left(1-\frac{1}{10}\right)=10-\frac{9}{10}=\frac{91}{10}\)
Tổng quát: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Ta có: \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
gọi hai số đó là a và b
ta có 7a.3b=21=> (a.b) .(7.3) =21=> a.b = 21:21=> a.b= 1 ma tất cả các số nhân với 1 sẽ bằng 1 nên sẽ không có tổng bình phương
TA CÓ:\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}\)
\(=1-\frac{1}{8}=\frac{7}{8}\)
A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 = 7/8