có 12 người đi du lịch tối đến mọi người muốn đi ngủ mà chỉ có 12 cái võng ở gần đó có 6 cái cột làm thế nào để xếp 12 cái võng lên 6 cái cột sao cho 12 cái võng không chồng đè hay chồng chéo lên nhau? 

#toán lớp 6

1: \(\dfrac{21}{-20}-\dfrac{11}{5}-\dfrac{-5}{4}\)

\(=-\dfrac{21}{20}-\dfrac{11}{5}+\dfrac{5}{4}\)

\(=-\dfrac{21}{20}-\dfrac{44}{20}+\dfrac{25}{20}=\dfrac{-40}{20}=-2\)

2: \(\dfrac{-1}{6}-\dfrac{4}{-3}-\dfrac{11}{18}\)

\(=\dfrac{-1}{6}+\dfrac{4}{3}-\dfrac{11}{18}\)

\(=-\dfrac{3}{18}+\dfrac{24}{18}-\dfrac{1}{18}=\dfrac{10}{18}=\dfrac{5}{9}\)

3: \(\dfrac{7}{17}-\dfrac{5}{-2}+\dfrac{13}{34}\)

\(=\dfrac{7}{17}+\dfrac{5}{2}+\dfrac{13}{34}\)

\(=\dfrac{14}{34}+\dfrac{85}{34}+\dfrac{13}{34}=\dfrac{112}{34}=\dfrac{56}{17}\)

4: \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)

\(=\left(\dfrac{15}{145}-\dfrac{29}{145}\right)\cdot\dfrac{29}{3}\)

\(=\dfrac{-14}{145}\cdot\dfrac{29}{3}=-\dfrac{14}{3}\cdot\dfrac{29}{145}=\dfrac{-14}{15}\)

5: \(\dfrac{3}{4}:\dfrac{7}{8}-3\dfrac{1}{4}+1\)

\(=\dfrac{3}{4}\cdot\dfrac{8}{7}-\dfrac{13}{4}+1\)

\(=\dfrac{6}{7}-\dfrac{9}{4}\)

\(=\dfrac{24}{28}-\dfrac{54}{28}=\dfrac{-30}{28}=-\dfrac{15}{14}\)

6: \(\left(0,75+\dfrac{-1}{3}-\dfrac{5}{8}\right):\dfrac{-5}{6}\)

\(=\left(\dfrac{3}{4}-\dfrac{1}{3}-\dfrac{5}{8}\right):\dfrac{-5}{6}\)

\(=\left(\dfrac{18}{24}-\dfrac{8}{24}-\dfrac{15}{24}\right):\dfrac{-5}{6}\)

\(=\dfrac{-5}{24}\cdot\dfrac{6}{-5}=\dfrac{6}{24}=\dfrac{1}{4}\)

7: \(3\dfrac{1}{7}-\dfrac{1}{4}\cdot\dfrac{2}{9}+25\%\)

\(=\dfrac{22}{7}-\dfrac{1}{18}+\dfrac{1}{4}\)

\(=\dfrac{841}{252}\)

8: \(4\dfrac{1}{5}-\dfrac{1}{5}\cdot\dfrac{10}{11}+15\%\)

\(=\dfrac{21}{5}-\dfrac{2}{11}+\dfrac{3}{20}\)

\(=\dfrac{924}{220}-\dfrac{40}{220}+\dfrac{66}{220}=\dfrac{950}{220}=\dfrac{95}{22}\)

9: \(1\dfrac{1}{2}-1\dfrac{2}{5}\cdot1,5-2,2:\dfrac{11}{4}\)

\(=\dfrac{3}{2}-\dfrac{7}{5}\cdot\dfrac{3}{2}-2,2\cdot\dfrac{4}{11}\)

\(=\dfrac{3}{2}\left(1-\dfrac{7}{5}\right)-\dfrac{8.8}{11}\)

\(=\dfrac{3}{2}\cdot\dfrac{-2}{5}-\dfrac{4}{5}=-\dfrac{3}{5}-\dfrac{4}{5}=-\dfrac{7}{5}\)

10: \(\dfrac{3}{17}\cdot\dfrac{13}{15}+\dfrac{3}{17}\cdot\dfrac{2}{15}\)

\(=\dfrac{3}{17}\left(\dfrac{13}{15}+\dfrac{2}{15}\right)=\dfrac{3}{17}\cdot\dfrac{15}{15}=\dfrac{3}{17}\)

11: \(\dfrac{7}{19}\cdot\dfrac{8}{11}+\dfrac{7}{19}\cdot\dfrac{3}{11}+\dfrac{-14}{19}\)

\(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{14}{19}=\dfrac{7}{19}-\dfrac{14}{19}=-\dfrac{7}{19}\)

12: \(\dfrac{-4}{5}\cdot\dfrac{7}{11}+\dfrac{-4}{5}\cdot\dfrac{4}{11}+1\dfrac{2}{11}\)

\(=\dfrac{-4}{5}\left(\dfrac{7}{11}+\dfrac{4}{11}\right)+1+\dfrac{2}{11}=-\dfrac{4}{5}+1+\dfrac{2}{11}\)

\(=\dfrac{1}{5}+\dfrac{2}{11}=\dfrac{11}{55}+\dfrac{10}{55}=\dfrac{21}{55}\)

\(\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\dfrac{1}{2019\cdot2021}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2020^2-1}\right)\)

\(=\dfrac{2^2}{2^2-1}\cdot\dfrac{3^2}{3^2-1}\cdot...\cdot\dfrac{2020^2}{2020^2-1}\)

\(=\dfrac{2\cdot3\cdot...\cdot2020}{1\cdot2\cdot...\cdot2019}\cdot\dfrac{2\cdot3\cdot...\cdot2020}{3\cdot4\cdot...\cdot2021}\)

\(=\dfrac{2020}{1}\cdot\dfrac{2}{2021}=\dfrac{4040}{2021}\)

\(\Rightarrow3\left(a-b\right)=5\left(a-b\right)\) 

\(\Leftrightarrow2\left(a-b\right)=0\Leftrightarrow a-b=0\Leftrightarrow a=b\)

Từ

\(3\left(a-b\right)=\dfrac{a}{b}\Rightarrow\dfrac{a}{b}=0\Rightarrow a=0\)

\(\Rightarrow a=b=0\) mà \(b\ne0\)

=> Dãy đẳng thức trên không tồn tại

\(\dfrac{1}{3}\cdot x-0,5\cdot x=\dfrac{3}{4}\)

=>\(x\left(\dfrac{1}{3}-\dfrac{1}{2}\right)=\dfrac{3}{4}\)

=>\(x\cdot\dfrac{-1}{6}=\dfrac{3}{4}\)

=>\(x=-\dfrac{3}{4}:\dfrac{1}{6}=-\dfrac{3}{4}\cdot6=-\dfrac{18}{4}=-\dfrac{9}{2}\)

\(\dfrac{1}{3}\)* x - 0,5 * x = \(\dfrac{3}{4}\)

\(\dfrac{1}{3}\)* x - \(\dfrac{1}{2}\)* x   = \(\dfrac{3}{4}\)

x * ( \(\dfrac{1}{3}\)- \(\dfrac{1}{2}\))    = \(\dfrac{3}{4}\)

x* - \(\dfrac{1}{6}\)            = \(\dfrac{3}{4}\)

x                    = - \(\dfrac{9}{2}\)

\(\dfrac{3}{1.3}+\dfrac{3}{1.5}+...+\dfrac{3}{97.99}\)

\(=\dfrac{3}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+...+\dfrac{1}{97.99}\right)\)

\(=\dfrac{3}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}.\left(1-\dfrac{1}{99}\right)\)

\(=\dfrac{3}{2}.\dfrac{98}{99}\)

\(=\dfrac{1}{1}.\dfrac{49}{33}\)

\(=\dfrac{49}{33}\)

= 1 - 1/3 + 1/ 3 - 1/5 + 1/ 5 - ... + 1/97 -1/99

=1-1/99

=98/99

k: \(\dfrac{3}{4}+\dfrac{1}{4}\cdot\left(1,25-\dfrac{3}{4}\right)\cdot\dfrac{5}{8}\)

\(=\dfrac{3}{4}+\dfrac{1}{4}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}\)

\(=\dfrac{3}{4}+\dfrac{5}{64}=\dfrac{48}{64}+\dfrac{5}{64}=\dfrac{53}{64}\)

m: \(25\%-1\dfrac{1}{2}+0,5\cdot\dfrac{12}{5}\)

\(=0,25-1,5+1,2\)

=-1,25+1,2

=-0,05

n: \(6\dfrac{9}{10}+\left(\dfrac{2}{5}-\dfrac{1}{10}\right)\cdot50\%\)

\(=6,9+0,5\cdot\left(0,4-0,1\right)\)

\(=6,9+0,5\cdot0,3=6,9+0,15=7,05\)

                                             Giải

a] A= SSH [1000 -1] .1 +1 = 1000 [số số hạng]

Tổng [1000 +1].1000 chia 2 = 500500

B= 1.2.3...10.11 = 39916800

Vì 500500 nhỏ hơn 39916800

Nên A bé hơn B