Tính hợp lý
A= 1/15+1/35+1/63+1/99+.....+1/129.15
gọi ý:15=3*5;35=5*7;63=7*9;...;129.15=43*45
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LT
3
11 tháng 4 2016
A=1/1.3 + 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13 + 1/13.15
A=1/1 - 1/3 +1/3 - 1/5 +1/5 -1/7+......+1/13 - 1/15
A=1 - 1/15
A=1/14
PM
0
25 tháng 7 2019
Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)
\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
\(A=\frac{5}{39}\)
Câu còn lại cx dựa như vậy nhé bn !
Chúc bn hc tốt <3
VH
3
BD
23 tháng 8 2023
\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)
\(=\dfrac{1}{3}+\dfrac{1}{25}\)
\(=\dfrac{28}{75}\)
S
15 tháng 4 2019
\(\frac{3x}{5}=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
Ta có: \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{13}\right)\)
\(=\frac{1}{2}.\frac{12}{13}\)
\(=\frac{6}{13}\)
\(\frac{3x}{5}=\frac{6}{13}\)
\(\Rightarrow3x=\frac{6.5}{13}\)
\(\Rightarrow3x=\frac{30}{13}\)
\(\Rightarrow x=\frac{10}{13}\)
~Học tốt~
10 tháng 1 2016
a)Ta có:
A= 1/15+1/35+1/63+1/99+1/143
A= 1/3.5+1/5.7+1/7.9+1/9.11+1/11.13
2A= 2/3.5+2/5.7+2/7.9+2/9.11+2/11.13
2A= 1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13
Đơn giản đi ta được:
2A= 1/3-1/13
2A= 10/39
A= 5/39
Vậy A= 5/39
30 tháng 7 2016
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\)
\(A=\frac{1}{3}-\frac{1}{101}\)
\(A=\frac{98}{303}\)
30 tháng 7 2016
A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)
= \(\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{99.101}\right):2\)\(=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)
= \(\left(\frac{1}{3}-\frac{1}{101}\right):2=\frac{101-3}{303}:2=\frac{98}{303}:2=\frac{49}{303}\)
Dấu chấm trong bài là dấu nhân nha !
15 tháng 7 2015
Gọi dãy là A ta có :
A = 1/3.5 + 1/5.7 + 1/7.9 + 1/9.11 + 1/11.13
A = 1/2 . ( 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 )
A = 1/2 . ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 )
A = 1/2 . ( 1/3 - 1/13 )
A = 1/2 . 10/39
A = 5/39
HT
15 tháng 7 2015
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
=\(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
=\(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)
=\(\frac{1}{2}.\frac{10}{39}\)
=\(\frac{5}{39}\)
15 tháng 7 2015
1/15+1/35+1/63+1/99+1/143
=1/2x(1/15+1/35+1/63+1/99+1/143)
=1/2x(2/3x5+2/5x7+2/7x9+2/9x11+2/11x13)
=1/2x(1/3-1/5+1/7-1/9+1/9-1/11+1/11-1/13)
=1/2x(1/3-1/13)
=1/2x10/39
=5/39
A = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{129.15}\)
A = \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{129.15}\)
A = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{129}-\frac{1}{15}\)
A = \(\frac{1}{3}-\frac{1}{15}\)= \(\frac{5}{15}-\frac{1}{15}\)
A = \(\frac{4}{15}\)
CẢM ƠN LỜI GỢI Ý
kết quả đúng là 7/45 nhé