Ta có: \(P=1\frac13\times1\frac18\times1\frac{1}{15}\times\ldots\times1\frac{1}{80}\times1\frac{1}{99}\)

\(=\frac43\times\frac98\times\frac{16}{15}\times\ldots\times\frac{81}{80}\times\frac{100}{99}\)

\(=\frac{2\times2}{1\times3}\times\frac{3\times3}{2\times4}\times\ldots\times\frac{10\times10}{9\times11}\)

\(=\frac{2\times3\times\ldots\times10}{1\times2\times\ldots\times9}\times\frac{2\times3\times\ldots\times10}{3\times4\times\ldots\times11}=\frac{10}{1}\times\frac{2}{11}=\frac{20}{11}\)

1/3x1/8x...x1/99

=1/(1x3)x1/(2x4)x...x1/(9x11)

=1/(1x3x2x4x...x9x11)

=1/(1x2x3x3x4x4x5x5x...x9x9x10x11)

B=3/2x4/3x...........x2018/2017

=3x4x5x...........x2018/2x3x2x2x............x2017

=2x2018

=4036

A,C tương tự

\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot\dfrac{16}{15}\cdot\dfrac{25}{24}\cdot\dfrac{36}{35}=\dfrac{12}{7}\)

= 4/3*9/8*16/15*25/24*36/35

=2*2/1*3 * 3*3/2*4 *4*4/3*5 *5*5/4*6 * 6*6/5*7

= (2*3*4*5*6 / 1*2*3*4*5) * ( 2*3*4*5*6 / 3*4*5*6*7)

=6/1* 2/7 

= 12/7

a) \(A=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.10}+\dfrac{1}{143}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)+\dfrac{1}{143}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{100}\right)+\dfrac{1}{143}=\dfrac{1}{2}.\dfrac{99}{100}+\dfrac{1}{143}=\dfrac{99}{200}+\dfrac{1}{143}=\dfrac{99.143+200.1}{200.143}=\dfrac{14157+200}{28600}=\dfrac{14357}{28600}\)

b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=14950\)

\(\Rightarrow x+x+...+x+\left(1+2+...+99\right)=14950\)

\(\Rightarrow100x+\left(\left(99+1\right):2\right).99:2=14950\)

\(\Rightarrow100x+2475=14950\Rightarrow100x=12475\Rightarrow x=\dfrac{12475}{100}=\dfrac{499}{4}\)

tính tích của 10 hỗn số đầu tiên trong dãy các hỗn số sau : 

1 và 1/3 x 1 và 1/8 x 1 và 1/15 x 1 và 1/24 x 1 và 1/35 x .......

a:Sửa đề: 1x3+2x4+...+99x101

\(=1\times\left(1+2\right)+2\times\left(2+2\right)+\cdots+99\times\left(99+2\right)\)

\(=\left(1\times1+2\times2+\cdots+99\times99\right)+2\times\left(1+2+\cdots+99\right)\)

\(=\frac{99\times\left(99+1\right)\times\left(2\times99+1\right)}{6}+2\times\frac{99\times100}{2}\)

\(=\frac{99\times100\times199}{6}+99\times100=33\times50\times199+99\times100\)

\(=33\times50\times\left(199+3\times2\right)=33\times50\times205=338250\)

b: \(\frac89\times\frac{15}{16}\times\ldots\times\frac{2499}{2500}\)

\(=\left(1-\frac19\right)\times\left(1-\frac{1}{16}\right)\times\ldots\times\left(1-\frac{1}{2500}\right)\)

\(=\left(1-\frac13\right)\times\left(1-\frac14\right)\times\ldots\times\left(1-\frac{1}{50}\right)\times\left(1+\frac13\right)\times\left(1+\frac14\right)\times\ldots\times\left(1+\frac{1}{50}\right)\)

\(=\frac23\times\frac34\times\ldots\times\frac{49}{50}\times\frac43\times\frac54\times\ldots\times\frac{51}{50}=\frac{2}{50}\times\frac{51}{3}=\frac{17}{25}\)