Bài 1: 

\(=\dfrac{-3-39}{42}+\dfrac{-6-11}{17}-\dfrac{1}{6}=-\dfrac{119}{48}\)

Bài 2: 

=>x:5=-13/20

hay x=-65/20=-13/4

x/30 + 4/5= 79/30

x/30 = 79/30 - 4/5

x/30= 11/6

11/6= 55/30

=> x= 55

\(\dfrac{x}{5}=\dfrac{5}{6}-\dfrac{19}{30}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{25}{30}-\dfrac{19}{30}=\dfrac{6}{30}=\dfrac{1}{5}\)

\(\Leftrightarrow x=1\)

Ta có : \(\dfrac{x}{5}=\dfrac{5}{6}+\left(-\dfrac{19}{30}\right)=\dfrac{25}{30}-\dfrac{19}{30}=\dfrac{1}{5}\)

\(\Leftrightarrow5x=5\)

\(\Leftrightarrow x=1\)

Vậy ...

Ta có: \(\dfrac{x}{5}=\dfrac{5}{6}-\dfrac{19}{30}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{25}{30}-\dfrac{19}{30}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{6}{30}=\dfrac{1}{5}\)

hay x=1

Vậy: x=1

\(x\times\dfrac{4}{5}=3-\dfrac{2}{5}\)

\(x\times\dfrac{4}{5}=\dfrac{13}{5}\)

\(x=\dfrac{13}{5}:\dfrac{4}{5}\)

\(x=\dfrac{13}{5}\times\dfrac{5}{4}\)

\(x=\dfrac{13}{4}\)

Chiều dài HCN là :

  \(\dfrac{5}{6}:\dfrac{2}{3}=\dfrac{5}{4}\left(m\right)\)

Chu vi HCN là :

  \(\left(\dfrac{5}{4}+\dfrac{2}{3}\right)\times2=\dfrac{23}{6}\left(m\right)\)

     Đ/s :...

\(x\times\dfrac{4}{5}=\dfrac{13}{5}\\ x=\dfrac{13}{5}\div\dfrac{4}{5}\\ x=\dfrac{13}{4}\)

\(CD=\dfrac{5}{6}\div\dfrac{2}{3}=\dfrac{5}{4}\\ P=\left(\dfrac{5}{4}+\dfrac{2}{3}\right)\times2=\dfrac{23}{6}\)

x = 4/5 : 3/4

x = 16/15

x = 5/9 : 8/3

x = 5/9 x 3/8

x = 5/24

\(x=\dfrac{4}{5}\times\dfrac{4}{3}\)

\(x=\dfrac{16}{15}\)

-----------------------

\(x=\dfrac{5}{9}\times\dfrac{3}{8}\)

\(x=\dfrac{5}{24}\)

\(x\times\dfrac{3}{4}=\dfrac{4}{5}\)

\(x=\dfrac{4}{5}:\dfrac{3}{4}\)

\(x=\dfrac{4}{5}\times\dfrac{4}{3}\)

\(x=\dfrac{16}{15}\)

\(\dfrac{5}{9}:x=\dfrac{8}{3}\)

      \(x=\dfrac{5}{9}:\dfrac{8}{3}\)

      \(x=\dfrac{5}{9}\times\dfrac{3}{8}\)

     \(x=\dfrac{15}{72}\)

    \(x=\dfrac{5}{24}\)

X=4/15:2/5=2/3

2/3

X = 4/5 : 2/15

X = 2/3

\(x-\dfrac{1}{4}=\dfrac{-21}{10}\)

\(x=\dfrac{-21}{10}+\dfrac{1}{4}\)

\(x=\dfrac{-42}{20}+\dfrac{5}{20}=\dfrac{-37}{20}\)

\(x^2+x=12\)

\(\Rightarrow x^2+x-12=0\)

\(\Rightarrow x\left(x-3\right)+4\left(x-3\right)=0\Rightarrow\left(x-3\right)\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)