a) Ta chia hình vuông thành 4 phần bằng nhau, sau đó tô 1 phần:

b) Ta chia hình chữ nhật ra 3 phần bằng nhau, sau đó tô 2 phần:

1/4

2/3

182.(\(\frac{1+\frac13+\frac19+\frac{1}{27}}{2+\frac23+\frac29+\frac{2}{27}}\) : \(\frac{4-\frac47+\frac{4}{49}-\frac{4}{343}}{1-\frac17+\frac{1}{49}-\frac{1}{343}}\)) : \(\frac{919191}{808080}\)

= 182.(\(\frac{1+\frac13+\frac19+\frac{1}{27}}{2.\left(\frac13+\frac19+\frac{1}{27}\right)}\) : \(\frac{4.\left(1-\frac17+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac17+\frac{1}{49}-\frac{1}{343}}\)) : \(\frac{91}{80}\)

= 182.(\(\frac12\) : 4) x \(\frac{80}{91}\)

= 182.\(\frac18\) x \(\frac{80}{91}\)

=\(\frac{91.2.8.10}{8.91}\)

= 2.10

= 20

=182.\(\orbr{\begin{cases}1.\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)\\2.\left(\frac{1}{2}+\frac{1}{9}+\frac{1}{27}\right)\end{cases}}:\frac{4.\left(\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1.\left(\frac{1}{3}+\frac{1}{49}-\frac{1}{343}\right)}:\frac{91}{80} \)

=.\(182.\left(\frac{1}{2}:\frac{4}{1}\right).\frac{91}{80}\)

=\(182.\frac{1}{8}.\frac{91}{80}\)

=.\(182.\frac{91}{640}\)

=\(\frac{8281}{320}\)

\(=182.\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2.\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}:\frac{4.\left(1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{9}-\frac{1}{343}}\right]:\frac{919191}{808080}\)

\(=182.\frac{1}{8}.\frac{808080}{919191}=\frac{182}{8}.\frac{80}{91}=20\)

\(=\dfrac{29}{22}+\dfrac{41}{44}+\dfrac{33}{52}+\dfrac{73}{182}\)

\(=\dfrac{58+41}{44}+\dfrac{33}{52}+\dfrac{73}{182}\)

\(=\dfrac{9}{4}+\dfrac{33}{52}+\dfrac{73}{182}=\dfrac{23}{7}\)

a) \(x-\frac{5}{7}=\frac{1}{9}\Rightarrow x=\frac{1}{9}+\frac{5}{7}\Rightarrow x=\frac{52}{63}\)

b) \(\frac{-3}{7}-x=\frac{4}{5}+\frac{-2}{3}\Rightarrow\frac{-3}{7}-x=\frac{2}{15}\Rightarrow x=\frac{-3}{7}-\frac{2}{15}\Rightarrow x=\frac{-59}{105}\)

c) \(x-\frac{1}{5}=\frac{2}{7}.\frac{-11}{5}\Rightarrow x-\frac{1}{5}=\frac{-22}{35}\Rightarrow x=\frac{-22}{35}+\frac{1}{5}\Rightarrow x=\frac{-3}{7}\)

d) \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Rightarrow\frac{x}{182}=\frac{-15}{91}\Rightarrow x=\frac{\left(-15\right).182}{91}\Rightarrow x=-30\)

a, 30/48=5/8

b,-104/182=-4/7

hoc tot

...

a>\(\frac{30}{48}=\frac{5}{8}\)

b>\(\frac{-104}{182}=\frac{-4}{7}\)

\(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\)

\(\frac{x}{15}=\frac{-1}{15}\)

=> \(x=-1\)

\(\frac{x}{182}=\frac{-6}{14}\cdot\frac{35}{91}\)

\(\frac{x}{182}=\frac{-15}{91}\)

=> \(91x=182\cdot\left(-15\right)\)

=> \(91x=-2730\)

=> \(x=-30\)

g, \(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\Leftrightarrow\frac{x}{15}=\frac{3}{5}-\frac{2}{3}\Leftrightarrow\frac{x}{15}=-\frac{1}{15}\)

\(\Leftrightarrow x=-1\)

h, \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Leftrightarrow\frac{x}{182}=-\frac{15}{91}\Leftrightarrow\frac{x}{182}=\frac{-30}{182}\)

\(\Leftrightarrow x=-30\)

Các số vừa chia hết cho 2 vừa chia hết cho 5 là cá số có chữ số tận cùng là 0.

Tìm tập hợp các số tự nhiên $n$ vừa chia hết cho 2, vừa chia hết cho 5 và $136<n<182$

{ 140 ; 150 ; 160 ; 170 ; 180 }

\(n\in\left\{140;150;160;170;180\right\}\)