S=1.4+4.7+7.10+10.13+...+61.64
A = 15/1.4 + 15/4.7 + 15/7.10+...+15/61.64 =?
A = \(\dfrac{15}{1.4}\) + \(\dfrac{15}{4.7}\) + \(\dfrac{15}{7.10}\) + ... + \(\dfrac{15}{61.64}\)
A = \(5.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}...+\dfrac{3}{61.64}\right)\)
A = 5.( \(\dfrac{1}{1}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{61}\) - \(\dfrac{1}{64}\))
A = 5.( \(\dfrac{1}{1}\) - \(\dfrac{1}{64}\))
A = 5. \(\dfrac{63}{64}\)
A = \(\dfrac{315}{64}\)
C=2/1.4+2/4.7+2/7.10+.......+2/61.64
\(C=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+........+\frac{3}{61.64}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{61}-\frac{1}{64}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{64}\right)\)
\(=\frac{2}{3}.\frac{63}{64}\)
\(=\frac{21}{32}\)
\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{61.61}\)
\(=2.\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{61.64}\right)\)
\(=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{61}-\frac{1}{64}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{64}\right)\)
\(=\frac{2}{3}.\frac{63}{64}\)
\(=\frac{21}{32}\)
\(C=\frac{2}{1.4}+\frac{2}{4.7}+.....+\frac{2}{61.64}\)
\(\Rightarrow\frac{3}{2}C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{61}-\frac{1}{64}\)
\(\Rightarrow C=\left(1-\frac{1}{64}\right):\frac{3}{2}=\frac{21}{32}\)
M = \(\dfrac{-3}{1.4}-\dfrac{3}{4.7}-\dfrac{3}{7.10}-...-\dfrac{3}{61.64}\)
\(=-\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{61}-\dfrac{1}{64}\right)=-\dfrac{1}{63}\)
So sánh A với 1, biết A= 3/1.4+3/4.7+3/7.10+....+3/61.64+3/64.67
( 31/1.4= 31 trên 3.4)
\(A=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{61\cdot64}+\dfrac{3}{64\cdot67}\)
\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\)
\(A=1-\dfrac{1}{67}\) < 1
=> A<1
Ta có:
\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{61.64}+\dfrac{3}{64.67}\)
\(=3.\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\right)\)
\(=3.\left(1-\dfrac{1}{67}\right)\)
\(=3.\dfrac{66}{67}\)
\(=\dfrac{198}{67}\)
Vì \(\dfrac{198}{67}\) có tử lớn hơn mẫu nên \(\dfrac{198}{67}>1\)
Vậy \(A>1\)
sửa bài:
... \(=1-\dfrac{1}{67}\)
\(=\dfrac{66}{67}\)
Vì \(\dfrac{66}{67}\) có tử nhỏ hơn mẫu nên \(\dfrac{66}{67}< 1\)
Vậy \(A< 1\)
3/1.4+3/4.7+3/7.10+3/10.13
3/1.4+3/4.7+3/7.10+3/10.13
=1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13
=1 - 1/13
=12/13
#)Trả lời :
Gọi tổng trên là A
\(A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}\)
\(A=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+\frac{3}{10}-\frac{3}{13}\)
\(A=3-\frac{3}{13}\)
\(A=2\frac{10}{13}=\frac{36}{13}\)
#~Will~be~Pens~#
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}\)
\(=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+\frac{3}{7}-\frac{3}{10}+\frac{3}{10}-\frac{3}{13}\)
\(=3-\frac{3}{13}\)
\(=\frac{36}{13}\)
3/1.4+3/4.7+3/7.10+3/10.13+3/13.6
\(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}\)
\(=\dfrac{3\cdot1}{1\cdot4}+\dfrac{3\cdot1}{4\cdot7}+\dfrac{3\cdot1}{7\cdot10}+\dfrac{3\cdot1}{10\cdot13}+\dfrac{3\cdot3}{13\cdot16}\)
\(=3\cdot\left(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+\dfrac{1}{10\cdot13}+\dfrac{1}{13\cdot16}\right)\)
\(=3\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)
\(=3\cdot\left(1-\dfrac{1}{16}\right)\)
\(=3\cdot\left(\dfrac{16}{16}-\dfrac{1}{16}\right)\)
\(=3\cdot\dfrac{15}{16}\)
\(=\dfrac{45}{16}\)
3/1.4 + 3/4.7 + 3/7.10 + 3/10.13 + 3/13.16
3/1.4 + 3/4.7 + .. +3/13.16
= 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16
= 1/1 - 1/16
= 15/16
\(=\frac{15}{16}\)
đúng cho mk nha Minh Thư Nguyễn
Bài 1: Tính tổng S
S=1/1.4+1/4.7+1/7.10+1/10.13+1/13.16+1/16.19+1/19.22
Bài 1: Tính tổng S
\(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{19.22}\)
\(4S=\dfrac{4}{1.4}+\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{19.22}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{19}-\dfrac{1}{22}\)
\(=1-\dfrac{1}{22}\)
\(S=\dfrac{21}{22}.\dfrac{1}{4}=\dfrac{21}{88}\)
Ta có:
A = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/16.19
3A= 1/3.(3/1.4 + 3/4.7 + 3/7.10 + ... + 3/16.19)
= 1/3. (1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/16 - 1/19)
= 1/3.(1 - 1/19)
= 1/3. 18/19
= 6/19