bài toán cần giải đáp ngay

H

hiensau99

a, $B= (\frac{1}{4}-1).(\frac{1}{9}-1).(\frac{1}{16}-1)....(\frac{1}{100}-1)$

$= \frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}....\frac{-99}{10^2}$

$=- (\frac{3.8.15....99}{2^2.3^2...10^2})$

$=- (\frac{1.3.2.4.3.5....9.11}{2^2.3^2...10^2})$

$=- (\frac{(1.2.3.5....9).(3.4.5...11)}{(2.3...10).(2.3...10)})$

$=- \frac{1.11}{10.2}$

$=\frac{-11}{20}$

b, Số thứ 98 là $\frac{9801}{9800}$

$A= \frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{9801}{9800}$

$= \frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{99^2}{98.100}$


$= \frac{2^2.3^2.4^2...99^2}{1.3.2.4.3.5...98.100}$


$= \frac{(2.3.4....99).(2.3.4....99)}{(1.2.3....97.98).(3.4.5...99.100)}$


$= \frac{99.2}{1.100}$

$= \frac{99}{50}$
 
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