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TH
0
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) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)
\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)
\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)
\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)
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~
S
6 tháng 4 2018
\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+....+\frac{1}{9.11}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{9.11}\)(tắt 1 bước nha)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{9}-\frac{1}{11}\)
\(2A=\frac{1}{3}-\frac{1}{11}\)
\(2A=\frac{8}{33}\)
\(\Rightarrow A=\frac{4}{33}\)
Vậy A=_____________
10 tháng 4 2018
ta có
\(\frac{1}{15}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}\right)\)
\(\frac{1}{35}=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(\frac{1}{63}=\frac{1}{2}\left(\frac{1}{7}-\frac{1}{9}\right)\)
......................................
\(\frac{1}{143}=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{13}\right)\)
Cộng hết lại: \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{1}{2}.\frac{10}{39}=\frac{5}{39}\)
10 tháng 4 2018
B = \(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(\Rightarrow B=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(\Rightarrow B=\frac{1.2}{3.5.2}+\frac{1.2}{5.7.2}+\frac{1.2}{7.9.2}+\frac{1.2}{9.11.2}+\frac{1.2}{11.13.2}\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(\Rightarrow B=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{10}{39}\)
\(\Rightarrow B=\frac{5}{39}\)
Vậy \(B=\frac{5}{39}\)
18 tháng 10 2020
nhìn công thức đây này \(\sqrt[]{\sqrt{ }\hept{\begin{cases}\\\end{cases}}\frac{ }{ }^{ }_{ }\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}\hept{\begin{cases}\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}}\) xong rồi đó không cần cảm ơn
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
OB
0
NN
1
KV
9 tháng 9 2023
B = 1/4 + 1/15 + 1/35 + 1/63 + 1/99 + 1/143 + 1/195
= 1/4 + 1/(3.5) + 1/(5.7) + 1/(7.9) + 1/(9.11) + 1/(11.13) + 1/(13.15)
= 1/4 + 1/2.(1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + 1/13 - 1/15)
= 1/4 + 1/2.(1/3 - 1/15)
= 1/4 + 1/2 . 4/15
= 1/4 + 2/15
= 23/60
P=1/3+1/15+1/35+1/63+1/99
=1:3+1:15+1:35+1:63+1:99
=1:(3+15+35+63+99)
=1:215
=1/215
Vậy:P=1/215