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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
ND
Nguyễn Đức Trí
VIP
15 tháng 7 2023
a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)
\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)
\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)
\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)
\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)
DT
1
VB
1
AM
12 tháng 6 2015
a)\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{2}{15}\)
b)\(M=1+3+3^2+...+3^{25}=\frac{3^{26}-1}{3-1}=\frac{3^{26}-1}{2}
20 tháng 7 2018
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
TN
20 tháng 7 2018
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
4 tháng 9 2015
A = 1/2 + 1/3 + 1/6 + 1/12 + 1/15 + 1/20 + 1/30 + 1/35 + 1/42 + 1/56 + 1/63 + 1/72 + 1/99
= ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 ) + ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
= ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9 ) + ( 1/1.3 + 1/3.5 + ... + 1/9.11 )
= ( 1 - 1/2 + 1/2 - 1/3 + ... + 1/8 - 1/9 ) + 1/2 . ( 2/1.3 + 2/3.5 + ... + 2/9.11 )
= ( 1 - 1/9 ) + 1/2 . ( 1 - 1/3 + 1/3 - 1.5 + ... + 1/9 - 1/11 )
= 8/9 + 1/2 . ( 1 - 1/11 )
= 8/9 + 1/2 . 10/11
= 8/9 + 5/11
= 133/99
4 tháng 9 2015
A = 1/2 + 1/3 + 1/6 + 1/12 + 1/15 + 1/20 + 1/30 + 1/35 + 1/42 + 1/56 + 1/63 + 1/72 + 1/99
= ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 ) + ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
= ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9 ) + ( 1/1.3 + 1/3.5 + ... + 1/9.11 )
= ( 1 - 1/2 + 1/2 - 1/3 + ... + 1/8 - 1/9 ) + 1/2 . ( 2/1.3 + 2/3.5 + ... + 2/9.11 )
= ( 1 - 1/9 ) + 1/2 . ( 1 - 1/3 + 1/3 - 1.5 + ... + 1/9 - 1/11 )
= 8/9 + 1/2 . ( 1 - 1/11 )
= 8/9 + 1/2 . 10/11
= 8/9 + 5/11
= 133/99