\(a\)) \(\dfrac{2}{3}+\dfrac{5}{8}+\dfrac{7}{12}\)

\(=\dfrac{16}{24}+\dfrac{15}{24}+\dfrac{7}{12}\)

\(=\dfrac{31}{24}+\dfrac{7}{12}\)

\(=\dfrac{31}{24}+\dfrac{14}{24}\)

\(=\dfrac{45}{24}\)

\(=\dfrac{15}{8}\)

\(b\)) \(1-\dfrac{8}{9}+\dfrac{5}{6}\)

\(=\dfrac{9}{9}-\dfrac{8}{9}+\dfrac{5}{6}\)

\(=\dfrac{1}{9}+\dfrac{5}{6}\)

\(=\dfrac{2}{18}+\dfrac{15}{18}\)

\(=\dfrac{17}{18}\)

\(c\)) \(\dfrac{13}{12}-\dfrac{2}{3}-\dfrac{1}{4}\)

\(=\dfrac{13}{12}-\dfrac{8}{12}-\dfrac{1}{4}\)

\(=\dfrac{5}{12}-\dfrac{1}{4}\)

\(=\dfrac{5}{12}-\dfrac{3}{12}\)

\(=\dfrac{2}{12}\)

\(=\dfrac{1}{6}\)

          \(\dfrac{9}{4}m=\dfrac{90}{4}dm\)

     Chiều dài của HCN đó là :

          \(6:\dfrac{90}{4}=\dfrac{4}{15}\left(dm\right)\)

     Chu vi của HCN đó là :

          \(\left(\dfrac{4}{15}+\dfrac{90}{4}\right)\cdot2=\dfrac{683}{15}\left(dm\right)\)

                Đáp số \(:\) \(\dfrac{638}{15}\left(dm\right)\)

\(a\)) \(\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{1}{6}\cdot5\)

\(=\dfrac{2}{3}+\dfrac{1}{4}-\dfrac{5}{6}\)

\(=\dfrac{8}{12}+\dfrac{3}{12}-\dfrac{5}{6}\)

\(=\dfrac{11}{12}-\dfrac{5}{6}\)

\(=\dfrac{11}{12}-\dfrac{10}{12}\)

\(=\dfrac{1}{12}\)

\(b\))\(\dfrac{19}{33}+\dfrac{35}{48}:\dfrac{7}{16}\)

\(=\dfrac{19}{33}+\dfrac{35}{48}\cdot\dfrac{16}{7}\)

\(=\dfrac{19}{33}+\dfrac{560}{336}\)

\(=\dfrac{19}{33}+\dfrac{5}{3}\)

\(=\dfrac{19}{33}+\dfrac{55}{33}\)

\(=\dfrac{74}{33}\)

\(\dfrac{12}{19}\) x \(\dfrac{1}{6}+\dfrac{1}{3}\) x \(\dfrac{3}{4}\)

\(=\dfrac{12}{114}+\dfrac{3}{12}\)

\(=\dfrac{2}{19}+\dfrac{1}{4}\)

\(=\dfrac{8}{76}+\dfrac{19}{76}\)

\(=\dfrac{27}{76}\)

Số số hạng :

(\(50-1\)) \(:\) \(1+1\) = \(50\)

Tổng

*\(A=\) [ ( \(50+1\) ) \(.\) \(50\) ] \(:2\)

*\(A=51\) . \(50:2\)

*\(A=2550:2\)

*\(A=1275\)

\(A.5,01\)

a) \(x\) + \(\dfrac{3}{4}\) = \(\dfrac{4}{3}\)                            b) \(\dfrac{1}{8}\) : \(x\) = \(\dfrac{1}{2}\)

    \(x\) = \(\dfrac{4}{3}\) - \(\dfrac{3}{4}\)                                 \(x\) = \(\dfrac{1}{8}\) : \(\dfrac{1}{2}\)

    \(x\) = \(\dfrac{7}{12}\)                                      \(x\) = \(\dfrac{1}{4}\)

c) \(x\) + \(\dfrac{1}{4}\) x \(\dfrac{6}{7}\) = \(\dfrac{3}{5}\)                    d) \(x\) + \(\dfrac{1}{4}\) x \(\dfrac{6}{7}\) = \(\dfrac{3}{5}\)  

   \(x\) + \(\dfrac{3}{14}\) = \(\dfrac{3}{5}\)                              \(x\) + \(\dfrac{3}{14}\) = \(\dfrac{3}{5}\)

  \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{3}{14}\)                                \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{3}{14}\)

  \(x\) = \(\dfrac{27}{70}\)                                      \(x\) = \(\dfrac{27}{70}\)

e) \(\left(\dfrac{2}{5}+\dfrac{4}{7}\right)\) : \(x\) = \(\dfrac{17}{5}\)             f) \(\dfrac{7}{3}\) - \(x\) = \(\dfrac{11}{5}\) : \(\dfrac{5}{6}\)

      \(\dfrac{34}{35}\) : \(x\) = \(\dfrac{17}{5}\)                          \(\dfrac{7}{3}\) - \(x\) = \(\dfrac{66}{25}\)

      \(x\) = \(\dfrac{34}{35}\) : \(\dfrac{17}{5}\)                           \(x\) = \(\dfrac{7}{3}\) - \(\dfrac{66}{25}\)

      \(x\) = \(\dfrac{2}{7}\)                                     \(x\) = - \(\dfrac{23}{75}\)                     

Để : Chiếc điện thoại này em dùng để học trực tuyến

x×0,75+34×5+2:43=10,5

$\Rightarrow x\times 0,75+0,75\times 5+2\times \frac{3}{4}=10,5$

$\Rightarrow x\times 0,75+0,75\times 5+2\times 0,75=10,5$

$\Rightarrow 0,75\times (x+5+2)=10,5$

$\Rightarrow 0,75\times (x+7)=10,5$

$\Rightarrow x+7=10,5:0,75$

$\Rightarrow x+7=14$

$\Rightarrow x=14-7$

$\Rightarrow x=7$

Vậy $x=7$