Question

[Tran Thu]

giups con voi co Hoai oi

 

Answer 1

[Nguyễn Thị Thương Hoài]

Bài 1:

a; \(\dfrac{-24}{11}\) + \(\dfrac{-19}{13}\) - (\(\dfrac{-2}{11}\) + \(\dfrac{20}{13}\))

= - \(\dfrac{24}{11}\) - \(\dfrac{19}{13}\) + \(\dfrac{2}{11}\) - \(\dfrac{20}{13}\)

=  - (\(\dfrac{24}{11}\) - \(\dfrac{2}{11}\)) - (\(\dfrac{19}{13}\) + \(\dfrac{20}{13}\))

= - \(\dfrac{22}{11}\) - \(\dfrac{39}{13}\)

= - 2 - 3

= - 5

Answer 2

[Nguyễn Thị Thương Hoài]

Bài 6

a; A = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{5^2}\) + ... + \(\dfrac{1}{50^2}\)

       \(\dfrac{1}{3^2}\) = \(\dfrac{1}{9}\)

      \(\dfrac{1}{4^2}\) < \(\dfrac{1}{3.4}\) = \(\dfrac{1}{3}-\dfrac{1}{4}\)

      \(\dfrac{1}{5^2}\) < \(\dfrac{1}{4.5}\)  = \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)

.....................................

      \(\dfrac{1}{50^2}\) < \(\dfrac{1}{49.50}\) = \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)

    Cộng vế với vế ta có:

A = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{50^2}\) < \(\dfrac{1}{9}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{50}\) = \(\dfrac{4}{9}\) - \(\dfrac{1}{50}\) < \(\dfrac{4}{9}\) (1)

       \(\dfrac{1}{3^2}\) = \(\dfrac{1}{9}\)

        \(\dfrac{1}{4^2}\) > \(\dfrac{1}{4.5}\) = \(\dfrac{1}{4}-\dfrac{1}{5}\)

     ....................................

       \(\dfrac{1}{50^2}\) >  \(\dfrac{1}{49.50}\) = \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)

Cộng vế với vế ta có:

  A = \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\) + ... + \(\dfrac{1}{50^2}\) > \(\dfrac{1}{9}\)+ \(\dfrac{1}{4}\) -  \(\dfrac{1}{50}\) = \(\dfrac{1}{4}\) + (\(\dfrac{1}{9}\) - \(\dfrac{1}{50}\)) > \(\dfrac{1}{4}\) (2)

Kết hợp (1) và (2) ta có: \(\dfrac{1}{4}\) < A < \(\dfrac{4}{9}\) (đpcm)