Question

[Nguyễn Đỗ Dũng Giang]

Giúp mình với các bạn,mình cần gấp câu c1 và câu d2 ạ !

Mik cảm ơn trc!

Answer 1

[Nguyễn Thị Thương Hoài]

c;     C = \(\dfrac{28^{28}+28^{24}+...+28^4+1}{28^{30}+28^{28}+...+28^2+1}\)

        A =         1 + 28⁴ + 28⁸ + 28¹² + ...28²⁸

  28⁴A = 28⁴ + 28⁸ + 28¹² + ... + 28²⁸ + 28³²

28⁴A - A = 28⁴+ 28⁸+...+28²⁸+ 28³²- (1 + 28⁴ + 28⁸+...+28²⁸)

(28⁴ - 1)A = 28⁴ + 28⁸+ ...+ 28²⁸ + 28³² - 1 - 28⁴- ...- 28²⁸

(28⁴ - 1)A = (28³² - 1) + (28⁴ - 28⁴) + (28⁸ - 28⁸) + ... + (28²⁸ - 28²⁸)

(28⁴ - 1)A = 28³² - 1 + 0 + 0... + 0

            A = (28³² - 1): (28⁴ - 1)

  Đặt B = 28³⁰ + 28²⁸ + ... + 28² + 1

  28²B = 28³² + 28³⁰ + ... + 28⁴ + 28²

28²B - B = 28³² + 28³⁰ + ... + 28⁴ + 28² - (28³⁰ + 28²⁸ +...+1)

(28² - 1)B = 28³² + 28³⁰+...+28⁴ + 28² - 28³⁰ - 28²⁸ - ... 28²- 1

(28² - 1)B = (28³² - 1) + (28³⁰ - 28³⁰) +...+(28² - 28²)

(28² - 1)B = (28³² - 1) + 0 + 0 +...+ 0

(28² - 1)B = 28³² - 1 

             B = (28³² - 1): (28² - 1)

C = \(\dfrac{A}{B}\) = \(\dfrac{28^{32}-1}{28^4-1}\) : \(\dfrac{28^{32}-1}{28^2-1}\)

C = \(\dfrac{28^{32}-1}{28^4-1}\) \(\times\) \(\dfrac{28^2-1}{28^{32}-1}\)

C = \(\dfrac{28^2-1}{28^4-1}\)

C = \(\dfrac{1}{785}\) 

Answer 2

[Nguyễn Thị Thương Hoài]

                Câu d:

 \(\dfrac{x-1}{99}\) + \(\dfrac{x-2}{98}\) + \(\dfrac{x-3}{97}\) = \(\dfrac{x-4}{96}\) + \(\dfrac{x-5}{95}\) + \(\dfrac{x-6}{94}\)

(\(\dfrac{x-1}{99}\)-1)+(\(\dfrac{x-2}{98}\)-1)+(\(\dfrac{x-3}{97}\)-1) = (\(\dfrac{x-4}{96}\)-1) + (\(\dfrac{x-5}{95}\)-1)+(\(\dfrac{x-6}{94}\)-1)

\(\dfrac{x-100}{99}\)+\(\dfrac{x-100}{98}\)+\(\dfrac{x-100}{97}\) = \(\dfrac{x-100}{96}\)+\(\dfrac{x-100}{95}\)+\(\dfrac{x-100}{94}\)

\(\dfrac{x-100}{99}\)+\(\dfrac{x-100}{98}\)+\(\dfrac{x-100}{97}\)- \(\dfrac{x-100}{96}\)-\(\dfrac{x-100}{95}\)-\(\dfrac{x-100}{94}\) = 0

(\(x-100\)).(\(\dfrac{1}{99}\)+\(\dfrac{1}{98}\)+\(\dfrac{1}{97}\) - \(\dfrac{1}{96}\)-\(\dfrac{1}{95}\)-\(\dfrac{1}{94}\)) = 0

Vì\(\dfrac{1}{98}< \dfrac{1}{98}< \dfrac{1}{97}< \dfrac{1}{96}< \dfrac{1}{95}< \dfrac{1}{94}\)

Nên (\(\dfrac{1}{99}\) + \(\dfrac{1}{98}\) + \(\dfrac{1}{97}\) )- (\(\dfrac{1}{96}\) + \(\dfrac{1}{95}\) +\(\dfrac{1}{94}\) )< 0 

⇒\(x-100\) = 0

Vậy \(x\) = 100