Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : $16A=\dfrac{16}{6.10}+\dfrac{16}{7.9}+\dfrac{16}{8.8}+\dfrac{16}{9.7}+\dfrac{16}{10.6}$
$=>16A=\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{6}$
$=>16A=2.(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$
$=>A=\dfrac{1}{8}(dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10})$
\(A=\dfrac{1}{6.10}+\dfrac{1}{7.9}+\dfrac{1}{8.8}+\dfrac{1}{9.7}+\dfrac{1}{10.6}\)
\(16A=\dfrac{16}{6.10}+\dfrac{16}{7.9}+\dfrac{16}{8.8}+\dfrac{16}{9.7}+\dfrac{16}{10.6}\)
\(16A=\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{6}\)
\(16A=2\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)
\(A=2:16\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)
\(A=\dfrac{1}{8}\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\left(đpcm\right)\)
A = 1/1.2 + 1/3.4 + 1/5.6 + 1/7.8 + 1/9.10
A = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 - 1/10
A = ( 1 + 1/3 + 1/5 + 1/7 + 1/9) - ( 1/2 + 1/4 + 1/6 + 1/8 + 1/10)
A = ( 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10) - 2.( 1/2 + 1/4 + 1/6 + 1/8 + 1/10)
A = ( 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10) - ( 1 + 1/2 + 1/3 + 1/4 + 1/5)
A = 1/6 + 1/7 + 1/8 + 1/9 + 1/10
B = 1/6.10 + 1/7.9 + 1/8.8 + 1/9.7 + 1/10.6
16B = 16/6.10 + 16/7.9 + 16/8.8 + 16/9.7 + 16/10.6
16B = 1/6 + 1/10 + 1/7 + 1/9 + 1/8 + 1/8 + 1/9 + 1/7 + 1/10 + 1/6
16B = 2.( 1/6 + 1/7 + 1/8 + 1/9 + 1/10)
8B = 1/6 + 1/7 + 1/8 + 1/9 + 1/10
Ta có A = 8B
=> A : B = 8
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{2}-\frac{1}{10}\)
\(A=\frac{2}{5}\)