\(C=\left[1-\frac{1}{2}\right]\times\left[1-\frac{1}{3}\right]\times\left[1-\frac{1}{4}\right]\times...\times\left[1-\frac{1}{2018}\right]\)

   \(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2017}{2018}\)

\(=\frac{1\times2\times3\times...\times2017}{2\times3\times4\times...\times2018}\)

\(=\frac{1}{2018}\)

Học tốt # ^-<

\(B=45.\left(\frac{7373}{3030}+\frac{7373}{4242}+\frac{7373}{5656}+\frac{7373}{9090}\right)\)

\(B=45.\left(\frac{73}{30}+\frac{73}{42}+\frac{73}{56}+\frac{73}{90}\right)\)

\(B=45.73.\left(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{90}\right)\)

\(B=3285.\left(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{9.10}\right)\)

\(B=3285.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{90}\right)\)

\(B=3285.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{90}\right)\)

\(B=3285.\frac{31}{360}=\frac{2263}{8}\)

bn xem lại đề đi nhé

a) Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(2A=\left(\frac{1}{2}\times2\right)+\left(\frac{1}{4}\times2\right)+\left(\frac{1}{8}\times2\right)+\left(\frac{1}{16}\times2\right)+\left(\frac{1}{32}\times2\right)\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

Ta lấy : \(2A-1A=1A\)

\(A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)

\(A=1-\frac{1}{32}\)

\(A=\frac{31}{32}\)

Vậy \(A=\frac{31}{32}\)

b) Đặt \(B=\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{18\times19}+\frac{2}{19\times20}\)

\(B=2\times(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20})\)

\(B=2\times\left(1-\frac{1}{20}\right)\)

\(B=2\times\frac{19}{20}\)

\(B=\frac{19}{10}\)

Vậy \(B=\frac{19}{10}\)

Học tốt # ^-<

dau gach cheo la dau gach ngang

A : 5 phút 

Đáp án là A

ta có: 

\(abc5-1107=5abc\)

\(1107=abc5-5abc\)

\(\Rightarrow abc5-5abc=1107\)

\(\Rightarrow abc\cdot10+5-5000-abc=1107\)

\(abc\cdot10-abc=1107+5000-5\)

\(abc\cdot9=6102\)

\(abc=6102:9=678\)

VẬY \(abc=678\)

TK MK NHA. ^_^

678 nha kb nha

abc2- 2abc = 2142

abc x 10 + 2 - abc + 2000 = 2142

abc x 10 + 2 - abc = 2142 + 2000

abc x 10 + 2 - abc = 4142

abc x 9 = 4142 - 2

abc x 9 = 4140

abc = 4140 : 9

abc = 460

Vậy abc = 460

2-c=2 => c=0

ab02-2ab0=2142

0-b=4 (nhớ 1) => b=6

a602-2a60=2142

5-a=1 => a=4 

nên abc=460

1abcd x 5 =abcd5 
(10000+abcd).5=abcd.10+5 
50000+5.abcd=10.abcd+5 
5.abcd=50000-5 
abcd=49995:5 
abcd=9999

Mik ko bt đúng ko nhưng hok tốt !

Gọi số cần tìm là \(\overline{abcd}\)(a khác 0: a,b,c,d <10)

Ta có: \(\overline{abcd5}\)=5x \(\overline{1abcd}\)

    \(\overline{abcd}\)x 10 +5 = 5 x (10000 + \(\overline{abcd}\))

\(\overline{abcd}\) x 10 + 5 = 50000 + 5 x \(\overline{abcd}\)

\(\overline{abcd}\)x 10 -5 x  \(\overline{abcd}\)= 50000 - 5

\(\overline{abcd}\) x 5 = 49995

\(\overline{abcd}\)= 49995 : 5 = 9999

Vậy số cần tìm là 9999.