-(-30)-(-x)=30

30+x=30

      x=30-30

      x=0

vậy x=0

-(-30)-(-x)=13

30+x=13

=> x= 13 - 30

     x= -17

-(-85)-(-71)+15+(-85)

=85+71+15-85

=(85-85)+(71+15)

=0+86

=86

\(-\left(-85\right)-\left(-71\right)+15+\left(-85\right)\)

\(=85+71+15-85\)

\(=\left(85-85\right)+\left(71+15\right)\)

\(=0+86\)

\(=86\)

\(\frac{2019.2020-4040}{2017.2018+4034}\)=\(\frac{\left(2017+2\right).2020-4040}{2017.2018+2017.2}\)

=\(\frac{2017.2020+2.2020-4040}{2017.\left(2018+2\right)}\)

=\(\frac{2017.2020+4040-4040}{2017.2020}\)

=\(\frac{2017.2020+0}{2017.2020}\)

=\(\frac{1}{1}\)=1

Ta có :
\(\overline{abbc}=\overline{ab}\times\overline{ac}\times7\)( 1 ) 

\(\Leftrightarrow100\times\overline{ab}+\overline{bc}=7\times\overline{ab}\times\overline{ac}\)

\(\Leftrightarrow\overline{ab}\left(7\times\overline{ac}-100\right)=\overline{bc}\)

\(\Leftrightarrow7\times\overline{ac}-100=\frac{\overline{bc}}{\overline{ab}}\)

Vì \(0< \frac{\overline{bc}}{\overline{ab}}< 10\)

\(\Leftrightarrow0< 7\times\overline{ac}-1000< 10\)

\(\Leftrightarrow100< 7\times\overline{ac< 110}\)

\(\Leftrightarrow14< \frac{100}{7}< \overline{ac}< \frac{110}{7}< 16\)

\(\Rightarrow\overline{ac}=15\)

Thay vào \(\left(1\right)\)ta được :
\(\overline{1bb5}=1b\times15\times7\)

\(\Leftrightarrow1005+110b=1050+105b\)

\(\Leftrightarrow5b=45\Leftrightarrow b=9\)

Vậy \(\hept{\begin{cases}a=1\\b=9\\c=5\end{cases}}\)

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