Tính nhanh:
1. \(\dfrac{1}{15}\)+ \(\dfrac{1}{35}\)+ \(\dfrac{1}{63}\)+ \(\dfrac{1}{99}\)+ \(\dfrac{1}{143}\)
2. \(\dfrac{5}{1.3}\)+ \(\dfrac{5}{1.5}\)+ \(\dfrac{5}{5.7}\)+ \(\dfrac{5}{7.9}\)+ \(\dfrac{5}{9.11}\)
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a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
a)\(=\dfrac{211}{180}\)
b)\(=\dfrac{5}{39}\)
c)=\(=-\dfrac{65}{168}\)
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
I. Tính:
1) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=1-\dfrac{1}{6}\)
\(=\dfrac{5}{6}\)
2) \(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)
\(=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=\dfrac{1}{3}-\dfrac{1}{13}\)
\(=\dfrac{13}{39}-\dfrac{3}{39}=\dfrac{10}{39}\)
II. Tìm x:
\(1\dfrac{3}{5}+\left(\dfrac{\dfrac{2}{171}}{\dfrac{5}{171}}+\dfrac{\dfrac{2}{373}}{\dfrac{5}{373}}\right)x=\dfrac{16}{5}\)
\(\dfrac{8}{5}+\left[\dfrac{2\left(\dfrac{1}{171}+\dfrac{1}{373}\right)}{5\left(\dfrac{1}{171}+\dfrac{1}{373}\right)}\right]x=\dfrac{16}{5}\)
\(\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\)
\(\dfrac{2}{5}x=\dfrac{16}{5}-\dfrac{8}{5}\)
\(\dfrac{2}{5}x=\dfrac{8}{5}\)
\(x=\dfrac{8}{5}:\dfrac{2}{5}\)
\(x=4\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}\cdot\dfrac{10}{39}=\dfrac{5}{39}\)
1: \(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=1/2*10/39
=5/39
2: \(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)=\dfrac{5}{2}\cdot\dfrac{10}{11}=\dfrac{50}{22}=\dfrac{25}{11}\)