`M=[x-2]/[x-3]=[x-3+1]/[x-3]=1+1/[x-3]`

Để \(M \in Z=>1+1/[x-3] \in Z\)

      \(=>1/[x-3] \in Z\)

  \(=>x-3 \in Ư_{1}\)

Mà \(Ư_{1}=`{`\pm 1`}

`@x-3=1=>x=4` (t/m)

`@x-3=-1=>x=2` (t/m)

Sfbvdfgvdaryyzg

3^100 = (3^10)^10

100^3 = 10000

3^10 = 59049

=> 3^100 > 100^3

A = 2008²⁰⁰⁸ + \(\dfrac{1}{2008^{2009}}\) + 1 

B = 2008²⁰⁰⁷ + \(\dfrac{1}{2008^{2008}}\) + 1

lấy vế trừ vế ta có:

A - B = 2008²⁰⁰⁸ -   2008²⁰⁰⁷ +   - \(\dfrac{1}{2008^{2008}}\)+\(\dfrac{1}{2008^{2009}}\)

A - B = 2008²⁰⁰⁷. ( 2008 - 1) +  - \(\dfrac{1}{2008^{2008}}\) + \(\dfrac{1}{2008^{2009}}\)

A - B = 2008²⁰⁰⁷. 2007  - \(\dfrac{1}{2008^{2008}}\) + \(\dfrac{1}{2008^{2009}}\)

2008²⁰⁰⁷.  2007 >   1 > \(\dfrac{1}{2008^{2008}}\) 

⇔ 2008²⁰⁰⁷.2007 - \(\dfrac{1}{2008^{2008}}\) > 0

⇔ 2008²⁰⁰⁷. 2007 - \(\dfrac{1}{2008^{2008}}\) + \(\dfrac{1}{2008^{2009}}\) > 0

⇔ A - B > 0 ⇔ A > B

kết luận A > B 

Ánh sáng đi từ Mặt Trời đến Trái Đất mất 151 giây nhé bạn

chúc bạn học tốt!!

a,    \(\dfrac{8}{7}\) x \(\dfrac{2}{3}\) - \(\dfrac{5}{6}\)x \(\dfrac{16}{14}\)

= \(\dfrac{16}{21}\) - \(\dfrac{20}{21}\)

= - \(\dfrac{4}{21}\) 

b,    x + \(\dfrac{3}{8}\) = \(\dfrac{12}{5}\) - \(\dfrac{3}{4}\)

      x  + \(\dfrac{3}{8}\) =   \(\dfrac{33}{20}\)

       x =      \(\dfrac{33}{20}\) - \(\dfrac{3}{8}\)

        x = \(\dfrac{51}{40}\)

c, x . \(\dfrac{8}{7}\) = \(\dfrac{4}{8}\) - \(\dfrac{3}{5}\) . \(\dfrac{2}{4}\)

   x. \(\dfrac{8}{7}\) = \(\dfrac{1}{2}\) - \(\dfrac{3}{10}\)

   x. \(\dfrac{8}{7}\) = \(\dfrac{1}{5}\)

   x = \(\dfrac{1}{5}\) : \(\dfrac{8}{7}\)

  x = \(\dfrac{1}{5}\) x \(\dfrac{7}{8}\)

 x = \(\dfrac{7}{40}\)

d, \(\dfrac{12}{5}\) - ( x + \(\dfrac{1}{3}\)) = \(\dfrac{3}{7}\) : \(\dfrac{4}{6}\)

   \(\dfrac{12}{5}\) - (x + \(\dfrac{1}{3}\)) = \(\dfrac{9}{14}\)

           x + \(\dfrac{1}{3}\) = \(\dfrac{12}{5}\) - \(\dfrac{9}{14}\) 

          x + \(\dfrac{1}{3}\)  = \(\dfrac{123}{70}\) 

         x = \(\dfrac{123}{70}\) - \(\dfrac{1}{3}\)

       x = \(\dfrac{299}{210}\)

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