a) $x-\genfrac{}{}{0ex}{}{2}{3}=\genfrac{}{}{0ex}{}{-5}{12}$

$x=\genfrac{}{}{0ex}{}{-5}{12}+\genfrac{}{}{0ex}{}{2}{3}$

$x=\genfrac{}{}{0ex}{}{-5}{12}+\genfrac{}{}{0ex}{}{8}{12}$

$x=\genfrac{}{}{0ex}{}{-5+8}{12}$

$x=\genfrac{}{}{0ex}{}{3}{12}$

$x=\genfrac{}{}{0ex}{}{1}{4}$

b) $\genfrac{}{}{0ex}{}{8}{5}:x=\genfrac{}{}{0ex}{}{-2}{3}$

$x=\genfrac{}{}{0ex}{}{8}{5}:\left(\genfrac{}{}{0ex}{}{-2}{3}\right)$

$x=\genfrac{}{}{0ex}{}{8}{5}.\left(\genfrac{}{}{0ex}{}{3}{-2}\right)$

$x=\genfrac{}{}{0ex}{}{-12}{5}$

c) $1-\genfrac{}{}{0ex}{}{3}{7}.x=-\genfrac{}{}{0ex}{}{2}{7}$

$\genfrac{}{}{0ex}{}{3}{7}.x=1-\left(-\genfrac{}{}{0ex}{}{2}{7}\right)$

$\genfrac{}{}{0ex}{}{3}{7}.x=\genfrac{}{}{0ex}{}{9}{7}$

$x=\genfrac{}{}{0ex}{}{9}{7}:\genfrac{}{}{0ex}{}{3}{7}$

$x=\genfrac{}{}{0ex}{}{9}{7}.\genfrac{}{}{0ex}{}{7}{3}$

$x=3$

a)

a) $\genfrac{}{}{0ex}{}{-2}{7}+\genfrac{}{}{0ex}{}{2}{7}:\genfrac{}{}{0ex}{}{3}{5}$

$=\genfrac{}{}{0ex}{}{-2}{7}+\genfrac{}{}{0ex}{}{2}{7}.\genfrac{}{}{0ex}{}{5}{3}$

$=\genfrac{}{}{0ex}{}{-2}{7}+\genfrac{}{}{0ex}{}{10}{21}$

$=\genfrac{}{}{0ex}{}{-6}{21}+\genfrac{}{}{0ex}{}{10}{21}$

$=\genfrac{}{}{0ex}{}{4}{21}$

b)$\genfrac{}{}{0ex}{}{-8}{19}+\genfrac{}{}{0ex}{}{-4}{21}-\genfrac{}{}{0ex}{}{17}{21}+\genfrac{}{}{0ex}{}{27}{19}$

$=\genfrac{}{}{0ex}{}{-8}{19}+\genfrac{}{}{0ex}{}{-4}{21}+\genfrac{}{}{0ex}{}{-17}{21}+\genfrac{}{}{0ex}{}{27}{19}$

$=\left(\genfrac{}{}{0ex}{}{-8}{19}+\genfrac{}{}{0ex}{}{27}{19}\right)+\left(\genfrac{}{}{0ex}{}{-4}{21}+\genfrac{}{}{0ex}{}{-17}{21}\right)$

$=\genfrac{}{}{0ex}{}{-8+27}{19}+\genfrac{}{}{0ex}{}{(-4)+(-17)}{21}$

$=\genfrac{}{}{0ex}{}{19}{19}+\genfrac{}{}{0ex}{}{-21}{21}$

$=1-1=0$

c) $\genfrac{}{}{0ex}{}{6}{5}.\genfrac{}{}{0ex}{}{3}{13}-\genfrac{}{}{0ex}{}{6}{5}.\genfrac{}{}{0ex}{}{16}{13}$

$=\genfrac{}{}{0ex}{}{6}{5}.\left(\genfrac{}{}{0ex}{}{3}{13}-\genfrac{}{}{0ex}{}{16}{13}\right)$

$=\genfrac{}{}{0ex}{}{6}{5}.\left(\genfrac{}{}{0ex}{}{3-16}{13}\right)$

$=\genfrac{}{}{0ex}{}{6}{5}.\left(\genfrac{}{}{0ex}{}{-13}{13}\right)$

$=\genfrac{}{}{0ex}{}{6}{5}.\left(-1\right)$

$=\genfrac{}{}{0ex}{}{-6}{5}.$

Chiều cao hình bình hành : 

       $24:3=8\left(m\right)$

Diện tích hình bình hành ban đầu :

        $20\times 8=160\left(m^2\right)$

Gọi số hàng có thể xếp được là : $x\left(x\in N^{\ast }\right)$

Theo bài ra : $x\in ƯC\left(48;18\right)$

Mà : $48=2^4.318=2.3^2$

$=>UCLN\left(48;18\right)=2.3=6$

$=>x\in Ư\left(6\right)=\left\{1;2;3;6\right\}$

Mà : $x\ge 5$

$=>x=6$

a) $(454-x)+4^3=116$

   454-x            = 116-4

  454-x             = 112

         x             =  454-112

        x              = 342

vậy x =342

b) $15$ chia hết cho $x+1$ với $x$ là số tự nhiên

Ư(15) = { 2;4;14}

a,−127+208−73+92

= (208+92)-[-208+(-73)

=300+208+(-73)

=508+(-73)

=435

b,2353−(473+2153)+(−55+373)

= 2353-2626+318

= -273+318

= 45

-20213<-20123