\(B2\\ a,MSC:60\\ \dfrac{5}{12}=\dfrac{5.5}{12.5}=\dfrac{25}{60};\dfrac{7}{20}=\dfrac{7.3}{20.3}=\dfrac{21}{60}\\ Vì:\dfrac{25}{60}>\dfrac{21}{60}.Nên:\dfrac{5}{12}>\dfrac{7}{20}\\ b,MSC:48\\ \dfrac{7}{12}=\dfrac{7.4}{12.4}=\dfrac{28}{48};\dfrac{3}{16}=\dfrac{3.3}{16.3}=\dfrac{9}{48}\\ Vì:\dfrac{28}{48}>\dfrac{9}{48}.Nên:\dfrac{7}{12}>\dfrac{3}{16}\\ c,\dfrac{4}{-10}=\dfrac{-2}{5}\\ MSC:25\\ \dfrac{4}{-10}=\dfrac{-2}{5}=\dfrac{-2.5}{5.5}=\dfrac{-10}{25}\\ Vì:\dfrac{-10}{25}< \dfrac{-3}{25}.Nên:\dfrac{4}{-10}< -\dfrac{3}{25}\\ d,\dfrac{-2}{30}=\dfrac{-1}{15}\\ MSC:210\\ \dfrac{-3}{14}=\dfrac{-3.15}{14.15}=\dfrac{-45}{210};\dfrac{-1}{15}=\dfrac{-1.14}{15.14}=\dfrac{-14}{210}\\ \dfrac{-45}{210}< \dfrac{-14}{210}.Nên:\dfrac{-3}{14}< \dfrac{-2}{30}\\ g,MSC:234\\ \dfrac{-7}{18}=\dfrac{-7.13}{18.13}=\dfrac{-91}{234};\dfrac{-3}{26}=\dfrac{-3.9}{26.9}=\dfrac{-27}{234}\\ Vì:\dfrac{-91}{234}< \dfrac{-27}{234}\Rightarrow\dfrac{-7}{18}< \dfrac{-3}{26}\)

Bài 3:

a: \(\dfrac{5}{22}+\dfrac{3}{22}=\dfrac{5+3}{22}=\dfrac{8}{22}=\dfrac{4}{11}\)

b: \(\dfrac{7}{12}-\dfrac{3}{20}\)

\(=\dfrac{35}{60}-\dfrac{9}{60}\)

\(=\dfrac{35-9}{60}=\dfrac{26}{60}=\dfrac{13}{30}\)

c: \(\dfrac{5}{18}+\dfrac{-3}{26}\)

\(=\dfrac{5}{18}-\dfrac{3}{26}\)

\(=\dfrac{5\cdot13-3\cdot9}{234}=\dfrac{65-27}{234}=\dfrac{38}{234}=\dfrac{19}{117}\)

d: \(\dfrac{7}{18}-\dfrac{3}{18}=\dfrac{7-3}{18}=\dfrac{4}{18}=\dfrac{2}{9}\)

b: \(-\dfrac{7}{14}-\dfrac{3}{-20}\)

\(=-\dfrac{1}{2}+\dfrac{3}{20}=\dfrac{-10+3}{20}=-\dfrac{7}{20}\)

c: \(\dfrac{-5}{24}-\dfrac{-3}{30}\)

\(=\dfrac{-5}{24}+\dfrac{1}{10}\)

\(=\dfrac{-25+12}{120}=-\dfrac{13}{120}\)

bài 1:

a: \(\dfrac{3}{15}=\dfrac{3\cdot4}{15\cdot4}=\dfrac{12}{60}\)

\(\dfrac{-2}{20}=\dfrac{-2\cdot3}{20\cdot3}=\dfrac{-6}{60}\)

b: \(\dfrac{1}{12}=\dfrac{1\cdot4}{12\cdot4}=\dfrac{4}{48}\)

\(\dfrac{5}{16}=\dfrac{5\cdot3}{16\cdot3}=\dfrac{15}{48}\)

c: \(\dfrac{4}{-15}=\dfrac{-4}{15}=\dfrac{-4\cdot5}{15\cdot5}=\dfrac{-20}{75}\)

\(\dfrac{-2}{25}=\dfrac{-2\cdot3}{25\cdot3}=\dfrac{-6}{75}\)

d: \(\dfrac{3}{14}=\dfrac{3\cdot15}{14\cdot15}=\dfrac{45}{210}\)

\(\dfrac{-2}{7}=\dfrac{-2\cdot30}{7\cdot30}=-\dfrac{60}{210}\)

\(\dfrac{1}{30}=\dfrac{1\cdot7}{30\cdot7}=\dfrac{7}{210}\)

g: \(\dfrac{-5}{18}=\dfrac{-5\cdot26}{18\cdot26}=\dfrac{-130}{468}\)

\(\dfrac{2}{12}=\dfrac{2\cdot39}{12\cdot39}=\dfrac{78}{468}\)

\(\dfrac{-5}{26}=\dfrac{-5\cdot18}{26\cdot18}=\dfrac{-90}{468}\)

90=90 mũ 1

90 = 90^1 

Vì p là số nguyên tố lớn hơn 3 nên  p là số lẻ

⇒ 17p là số lẻ ⇒ 17p + 1 là số chẵn là hợp số (trái với đề bài)

Vậy với p là số nguyên tố lớn hơn 3 thì việc chứng minh 17p + 1 là số nguyên tố là không thể.

2²⁰¹ = (2⁴)⁵⁰.2 = \(\overline{..6}\)⁵⁰.2 = \(\overline{..2}\)

2²⁰¹=2^4.50+1

=>2²⁰¹=....2

\(3^{x+1}\) + 3\(^{x+2}\) - 2 \(\times\) 3\(^x\) = 270

3\(^x\).(3 + 3² - 2) = 270

3\(^x\).10 = 270

3\(^x\)      = 270 : 10

3\(^x\)      = 27

3\(^x\)      = 3³

\(x\)       = 3

Vậy \(x\) = 3

3^x+1+3^x+2-2 . 3^x=270

3^x . 3+3^x . 3²-2 . 3^x=270

3^x . (3+9-2)          =270

3^x . 10                       =270

3^x                              =270 : 10 = 27 = 3³

=>x=3

\(\dfrac{3}{28}\) ≤   \(\dfrac{x}{56}\) ≤ \(\dfrac{1}{4}\)

\(\dfrac{6}{56}\) ≤    \(\dfrac{x}{56}\) ≤ \(\dfrac{14}{56}\)

6 ≤        \(x\)    ≤ 14

Vì \(x\) nguyên nên \(x\) \(\in\) {6; 7; 8; 9; 10; 11; 12; 13; 14}

Vậy \(x\) \(\in\) {6; 7; 8; 9; 10; 11; 12; 13; 14}

Việt Nam 2-4 Nhật Bản

Việt Nam 0-1 Indonesia

Việt Nam 2-3 Irap