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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HT
25 tháng 7 2015
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+.....+\frac{1}{129.15}=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+....+\frac{1}{1935}\)
=\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{43.45}\)
=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{43}-\frac{1}{45}\right)\)
=\(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{45}\right)\)
=\(\frac{1}{2}.\frac{43}{90}\)
=\(\frac{43}{180}\)
DP
31 tháng 5 2017
\(A=11+14+17+...+62+65\)
Số số hạng của \(A\)là
\(\left(65-11\right)\div3+1=19\)(số hạng)
Tổng của \(A\)là:
\(\left(11+65\right)\times19\div2=722\)
Đáp số: 722
\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(B=\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+.....+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\times\frac{1}{2}\)
\(B=\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{9}-\frac{1}{11}\right)\times\frac{1}{2}\)
\(B=\left(1-\frac{1}{11}\right)\times\frac{1}{2}\)
\(B=\frac{10}{11}\times\frac{1}{2}\)
\(B=\frac{5}{11}\)
\(C=\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+\frac{3}{154}\)
\(C=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
\(C=\left(\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+....+\frac{3}{11}-\frac{3}{14}\right)\div3\)
\(C=\left(\frac{3}{2}-\frac{3}{14}\right)\div3\)
\(C=\frac{9}{7}\div3\)
\(C=\frac{3}{7}\)
31 tháng 5 2017
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(B=1-\frac{1}{11}\)
\(B=\frac{10}{11}\)
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
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
WH
29 tháng 3 2018
Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)
\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)
Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)
\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)
\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)
\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)
\(\Rightarrow2B=1-\frac{1}{15}\)
\(\Rightarrow2B=\frac{14}{15}\)
\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)
\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)
30 tháng 4 2016
1) x - 2 = -6
x = - 6 + 2
x = -4
2) -5x-(-3)=13
-5x+3=13
-5x=13-3=10
x=10:(-5)
x=-2
3) 15-(x-7)=21
15-x+7=21
15-x=21-7=14
x=15-14=1
4) 3x+17=2
3x=2-17=-15
x=-15:3=-5
5) 45-(x-9)=-35
45-x+9=-35
45-x=-35-9=-44
x=45-(-44)=89
6) -5+x=15
x=15-(-5)
x=20
7) 2x-(-17)=15
2x+17=15
2x=15-17=-2
x=-2:2=-1
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