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a) C = 1/3 + 1/15 + ... + 1/2115
= 1/(1.3) + 1/(3.5) + ... + 1/(45.47)
= 1/2 . (1 - 1/3 + 1/3 - 1/5 + ... + 1/45 - 1/47}
= 1/2 . (1 - 1/47)
= 1/2 . 46/47
= 23/47
\(C=\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{2115}\\ \\ \\ \\ \\ C=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{45.47}\\ \\ \\ \\ \\ \Rightarrow2C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{45.47}\\ \\ \\ \\ \\ 2C=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{45}-\dfrac{1}{47}\\ \\ \\ \\ \\ 2C=\dfrac{1}{1}-\dfrac{1}{47}=\dfrac{46}{47}\\ \\ \\ \\ \\ \Rightarrow C=\dfrac{46}{47}:2\\ \\ \\ \\ \\ C=\dfrac{46}{47}\cdot\dfrac{1}{2}=\dfrac{23}{47}\)
<=> P = 2100 - ( 299 + 298 + ..... + 22 + 2 + 1 )
Đặt A = 1 + 2 + 22 + 23 + ...... + 298 + 299
<=> 2A = 2.( 1 + 2 + 22 + ..... + 298 + 299 )
<=> 2A = 2 + 22 + 23 + ...... + 299 + 2100
<=> 2A - A = ( 2 + 22 + 23 + ..... + 299 + 2100 ) - ( 1 + 2 + 22 + ..... + 298 + 299 )
<=> A = 2100 - 1
=> P = 2100 - ( 2100 - 1 )
=> P = - 1
Vậy P = - 1
=32.233+67.233+68.233
=233.(68+32+67)
=233.167
=38911
hok tốt k nhé bạn
\(32\times\left(233+67\right)+68\times233\)
=\(32\times233+67\times233+68\times233\)
=\(233\left(32+67+68\right)\)
=\(233\times167\)
=\(38911\)
#hoctot
15.37.4 + 120.21 + 21.5.12 = 15.4.37 + 21.2.60 + 5.12.21 = 60.37 + 42.60 + 60.21 = 60.(37 + 42 + 21) = 60.100 = 6000
b: \(=\dfrac{1}{9}\cdot\dfrac{3}{5}+\dfrac{5}{6}\cdot\dfrac{3}{5}-\dfrac{7}{2}\cdot\dfrac{3}{5}\)
\(=\dfrac{3}{5}\left(\dfrac{1}{9}+\dfrac{5}{6}-\dfrac{7}{2}\right)=\dfrac{3}{5}\cdot\dfrac{-23}{9}=\dfrac{-69}{45}=\dfrac{-23}{15}\)
d: \(=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(-\dfrac{20}{41}-\dfrac{21}{41}\right)-\dfrac{5}{7}=-\dfrac{5}{7}\)
819 + 364 + 181 + 636 - 636 - 364 - 100 + 32 : 4
= 819 + 181 + 364 - 364 + 636 - 636 - 100 + 8
= 1000 + 0 + 0 - 100 + 8
= 1000 - 100 + 8
= 900 + 8
= 908
c) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{1}{2}.\dfrac{10}{39}=\dfrac{5}{39}\)
e) \(\dfrac{1}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{1}{5}-\dfrac{1}{5}=\dfrac{1}{5}\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{1}{5}=\dfrac{1}{5}.1-\dfrac{1}{5}=0\)