\(E=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2006}\right)\left(1-\frac{1}{2007}\right)\)

\(E=\frac{1}{2}.\frac{2}{3}....\frac{2005}{2006}.\frac{2006}{2007}\)

\(E=\frac{1.2.3.4...2005.2006}{2.3.4.5....2006.2007}\)

\(E=\frac{1}{2007}\)

\(\frac{4.15.9.24}{3.12.8.5}\)

\(=\frac{1.5.3.3}{1.1.1.5}\)

\(=\frac{45}{5}=9\)

\(A=\frac{2}{1+2}+\frac{2+3}{1+2+3}+...+\frac{2+3+...+20}{1+2+3+...+20}\)

\(A=\frac{2}{3}+\frac{5}{6}+...+\frac{209}{210}\)

\(A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{210}\right)\)

\(A=\left(1+1+....+1\right)\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{210}\right)\)

\(A=19-\left(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{420}\right)\)

\(A=19-\left(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\right)\)

\(A=19-2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)

\(A=19-2\cdot\left(\frac{1}{2}-\frac{1}{21}\right)\)

\(A=19-2\cdot\frac{19}{42}=19-\frac{19}{21}=\frac{380}{21}\)

Vậy A= \(\frac{380}{21}\)

\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2005}\right)\left(1-\frac{1}{2006}\right)\)

\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2004}{2005}\cdot\frac{2005}{2006}\)

\(B=\frac{1\cdot2\cdot...\cdot2004\cdot2005}{2\cdot3\cdot...\cdot2005\cdot2006}\)

\(B=\frac{1}{2006}\)

Vậy \(B=\frac{1}{2006}\)

\(\dfrac{12}{6}-\dfrac{4}{6}-\dfrac{1}{6}x\dfrac{1}{2}=\dfrac{7}{6}x\dfrac{1}{2}=\dfrac{7}{12}\)

 bbgfhfygfdsdty64562gdfhgvfhgfhhhhh

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1/10 + 3/5 : 2/3

=1/10 + 9/10

=10/10

=1

\((\)6/8 -2/6\()\) x 1/2

=18/48 X 1/2

=3/7 x 1/2

=2/14

    \(\dfrac{1}{10}\) + \(\dfrac{3}{5}\):\(\dfrac{2}{3}\)

= \(\dfrac{1}{10}\) +  \(\dfrac{3}{5}\) \(\times\) \(\dfrac{3}{2}\)

= \(\dfrac{1}{10}\) + \(\dfrac{9}{10}\)

= \(\dfrac{10}{10}\)

= 1 

( \(\dfrac{6}{8}\) - \(\dfrac{2}{6}\)) \(\times\) \(\dfrac{1}{2}\)

=( \(\dfrac{18}{24}\)  - \(\dfrac{8}{24}\)) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{10}{24}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{5}{24}\)

Thanks kiuuuu

\(2\dfrac{1}{20}\)

\(\dfrac{13}{18}\)

A = 3 + 6 + 9 + ... + 2007 

=>A = 3( 1 + 2 + 3 + ... + 669 )

=> A = \(3\cdot\left(\frac{670\cdot669}{2}\right)\)

=> A = \(3\cdot224115\)= 672345

B = \(2\cdot53\cdot12+4\cdot6\cdot87-3\cdot8\cdot40\)

=> B = 24 * 53 + 24 * 87 - 24 * 40

=> B = 24 * ( 53 + 87 - 40 )

=> B = 24 * 100 = 2400

c) ta có Tử số = \(2006\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2007}\right)\)

Mẫu số = \(\frac{2007-1}{1}\)+\(\frac{2007-2}{2}\)+...+\(\frac{2007-2006}{2006}\)

=> Mẫu số = \(\frac{2007}{1}\)\(-1\)+ \(\frac{2007}{2}\)\(-1\)+ ... + \(\frac{2007}{2006}\)\(-1\)

=> Mẫu số = \(\frac{2007}{1}\)+ \(\frac{2007}{2}\)+ ... + \(\frac{2007}{2006}\)- ( 1 + 1 + 1 + ... + 1 )        ( 1 + 1 + ... + 1  có 2006 số hạng 1 )

=> Mẫu số =  ( 2007 - 2006 ) + \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}\right)\)

=> Mẫu số = \(\frac{2007}{2007}\)+ \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2006}\right)\)

=> Mẫu số = \(2007\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}\right)\)

=> C = \(\frac{TS}{MS}\)= \(\frac{2006}{2007}\)

=1/2*2/3*3/4*4/5*5/6=1/6

thank's nha