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p=1/(3*5)+1/(5*7)+.....+1/(2015*2017)+1/(2017*2019)
<=> p = 1/3-1/5+1/5-1/7+1/7-......+1/2017-1/2019
<=> p = 1/3 - 1/2019
<=> p = 224/673
\(P=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2019}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2017}-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2019}\right)\)
\(=\frac{112}{673}\)
5/14x7/13x26/25
Bạn để ý rút gọn 14 cho 7 như sau
5/2x1/13x26/25
=5/26x26/25
Rút gọn 26 với 26,ta đc 5/25=1/5
\(\frac{5}{14}\times\frac{7}{13}\times\frac{26}{25}\)\(=\frac{5\times7\times13\times2}{7\times2\times13\times5\times5}\)\(=\frac{1}{5}\)
a,\(\frac{31}{17}+\frac{5}{13}+\frac{8}{13}-\frac{14}{17}=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{5}{13}+\frac{8}{13}\right)=1+1=2\) b,\(\frac{5}{7}.\frac{22}{11}+\frac{5}{7}.\frac{9}{11}+\frac{5}{7}=\frac{5}{7}.\left(\frac{22}{11}+\frac{9}{11}+1\right)=\frac{5}{7}.4=\frac{20}{7}\)
Giải:
\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\)
\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\dfrac{47}{150}\)
\(B=\dfrac{47}{100}\)
Chúc em học tốt!
\(\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)
\(\Rightarrow c=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{97}-\frac{1}{99}\)
\(\Rightarrow C=\frac{1}{3}-\frac{1}{99}\)
\(\Rightarrow C=\frac{32}{99}\)
C = 2/3x5 + 2/5x7 + 2/7x9 + ... + 2/97x99
C = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99
C = 1/3 - 1/99
C = 32/99
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
\(\dfrac{33}{38}:\dfrac{11}{19}=\dfrac{33\times19}{38\times11}=\dfrac{3\times1}{2\times1}=\dfrac{3}{2}\)
\(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}\)
\(=6.\frac{1}{14}.\frac{1}{13}.\frac{9}{2}-\frac{1}{14}.\frac{1}{13}\)
\(=\frac{1}{14}.\frac{1}{13}.\left(\frac{6.9}{2}-1\right)\)
\(=\frac{1}{182}.26\)
\(=\frac{1}{7}\)
3/7.9/26-1/14.1/13=3.9/7.26-1/14.1/13
=27/182-1/14.1/13
=27/182-1.1/14.13
=27/182-1/182
=26/182=1/7