\(5.5.125^2=5.5.\left(5^3\right)^2=5.5.5^{3.2}=5^1.5^1.5^6\\ =5^{1+1+6}=5^8\)

5. 5. 125^2 

= 25 . 15625

= 390625

A = { 0;1;2;3;4;5}

A= { x\(\in\) \(ℕ\) | x \(\le\) 5}

\(a=\dfrac{5}{10.12}+\dfrac{5}{12.14}+\dfrac{5}{14.16}+...+\dfrac{5}{998.1000}\\ \dfrac{2}{5}a=\dfrac{2}{10.12}+\dfrac{2}{12.14}+\dfrac{2}{14.16}+...+\dfrac{2}{998.1000}\\ \dfrac{2}{5}a=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+...+\dfrac{1}{998}-\dfrac{1}{1000}\\ \dfrac{2}{5}a=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\\ a=\dfrac{99}{400}\)

\(a=\dfrac{5}{10.12}+\dfrac{5}{12.14}+...+\dfrac{5}{998.1000}\)

\(a=5\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+.....+\dfrac{1}{998}+\dfrac{1}{1000}\right)\)

\(A=5\left(\dfrac{1}{10}-\dfrac{1}{1000}\right)=\dfrac{99}{200}\)

17.85 + 15.17-120

= 17. ( 85 + 15) - 120

= 17 . 100 - 120

= 1700 - 120

= 1580

Gọi số bé là : \(\overline{ab}\)

=> Số lớn là : \(\overline{ab7}\)

Ta có :

\(\overline{ab7}-\overline{ab}=385\\ =>\overline{ab}x10+7-\overline{ab}=385\\ =>\overline{ab}x9=378\\ =>\overline{ab}=42;\overline{ab7}=427\)

21-(25-x+21)=-(25-7)

21-(25-x+21)=-18

      25-x+21 =21+18

      25-x+21 =39

      25-x       =39-21

      25-x       =18

           x       =25-18

           x       =7

vậy x=7

\(45:2\dfrac{4}{7}+50\%-1,25=45:\dfrac{18}{7}+\dfrac{50}{100}-\dfrac{125}{100}\\ =45.\dfrac{7}{18}+\dfrac{1}{2}-\dfrac{5}{4}\\ =\dfrac{35}{2}+\dfrac{1}{2}-\dfrac{5}{4}\\ =18-\dfrac{5}{4}=\dfrac{67}{4}\)

\(\dfrac{4}{5}+\dfrac{5}{7}.x=\dfrac{1}{6}\\ =>\dfrac{5}{7}.x=\dfrac{1}{6}-\dfrac{4}{5}=-\dfrac{19}{30}\\ =>x=-\dfrac{19}{30}:\dfrac{5}{7}\\ =>x=-\dfrac{133}{150}\)

TL:

S= \(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+....+ \(\dfrac{1}{89}\)+\(\dfrac{1}{90}\)

S=( \(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+...+\(\dfrac{1}{60}\))+(\(\dfrac{1}{61}\)+\(\dfrac{1}{62}\)+...+ \(\dfrac{1}{89}\)+\(\dfrac{1}{90}\))

Ta có:

( \(\dfrac{1}{31}\)+\(\dfrac{1}{32}\)+...+\(\dfrac{1}{60}\))> \(\dfrac{1}{60}\)+\(\dfrac{1}{60}\)+....+\(\dfrac{1}{60}\)=\(\dfrac{1}{2}\)

Và:

(\(\dfrac{1}{61}\)+\(\dfrac{1}{62}\)+...+ \(\dfrac{1}{89}\)+\(\dfrac{1}{90}\))>\(\dfrac{1}{90}\)+\(\dfrac{1}{90}\)+....+\(\dfrac{1}{90}\)=\(\dfrac{1}{3}\)

Vậy S> \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)=\(\dfrac{5}{6}\) (đpcm)