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26 tháng 3 2019

1, A=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)

   A= \(\frac{100}{2}\)

   A=50

2, B=\(\frac{-1}{2}.\frac{-2}{3}....\frac{-98}{99}\)

     B= \(\frac{1}{99}\)

26 tháng 3 2019

\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)

     \(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}......\frac{99}{98}\cdot\frac{100}{99}\)

     \(=\frac{100}{2}\)

       \(=50\)

\(B=\left(\frac{1}{2}-1\right)\cdot\left(\frac{1}{3}-1\right)\cdot\left(\frac{1}{4}-1\right)......\left(\frac{1}{99}-1\right)\)

     \(=\left(-\frac{1}{2}\right)\cdot\left(-\frac{2}{3}\right)\cdot\left(-\frac{3}{4}\right).....\left(-\frac{97}{98}\right)\cdot\left(-\frac{98}{99}\right)\)

       \(=-\frac{1}{99}\)

2 tháng 4 2023

1+1=3 :)))

29 tháng 3 2017

a) \(=\frac{3}{2}.\frac{4}{3}....\frac{100}{99}=\frac{100}{2}=50\)

29 tháng 3 2017

a) =3/2 . 4/3 . 5/4 ...100/99

   =\(\frac{3.4.5...100}{2.3.4..99}\)

  =\(\frac{100}{2}\)

b) =

25 tháng 3 2018

\(T=\left(\frac{1}{2}-\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{5}\right)\left(\frac{1}{2}-\frac{1}{7}\right).....\left(\frac{1}{2}-\frac{1}{99}\right)\)

\(T=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-........-\frac{1}{99}\right)\)

\(T=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-..........-\frac{1}{99}\right)\)

\(T=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(T=\frac{1}{2}\left(\frac{99}{297}-\frac{3}{297}\right)\)

\(T=\frac{1}{2}.\frac{96}{297}\)

\(T=\frac{1}{2}.\frac{32}{99}\)

\(T=\frac{16}{99}\)

23 tháng 4 2020

\(T=\left(\frac{1}{2}-\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{5}\right)\left(\frac{1}{2}-\frac{1}{7}\right)\cdot.....\cdot\left(\frac{1}{2}-\frac{1}{99}\right)\)

\(=\frac{1}{2\cdot3}\cdot\frac{3}{2\cdot5}\cdot\frac{5}{2\cdot7}\cdot.....\cdot\frac{97}{2\cdot99}\)

\(=\frac{1\cdot3\cdot5\cdot.....\cdot97}{2^{49}\left(3\cdot5\cdot7\cdot....\cdot99\right)}=\frac{1}{2^{49}\cdot99}\)

24 tháng 1 2018

\(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{98^2}-1\right)\left(\frac{1}{99^2}-1\right)\)

\(=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right).....\left(1-\frac{1}{98^2}\right)\left(1-\frac{1}{99^2}\right)\)

\(=\frac{3}{2^2}.\frac{8}{3^2}......\frac{9603}{98^2}.\frac{9800}{99^2}\)

\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{97.99}{98^2}.\frac{98.100}{99^2}\)

\(=\frac{1.2.4...97.98}{2.3....98.99}.\frac{3.4...99.100}{2.3....98.99}\)

\(=\frac{1}{99}.\frac{100}{2}\)

\(=\frac{50}{99}\)

24 tháng 1 2018

bn viết sai 1 chỗ nhưng ko s ^^ tks nhoa