\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)\(=\frac{7.512-5.1024}{\left(-256\right)}\)\(=\frac{3583-5120}{\left(-256\right)}\)\(=\frac{1537}{256}\)

Mình học hơi kém nếu sai mong mn đừng gạch đá .

1)

\(\frac{7.8^3-5.2^{10}}{\left(-16\right)^2}\)       

=  \(\frac{7.2^8.2-5.2^8.2^2}{16^2}\)

=  \(\frac{2^8.\left(2.7-5.2^2\right)}{2^8}\)

=   \(\frac{2^8.\left(-6\right)}{2^8}\)

=   \(-6\)

2)

b)\(\frac{-28}{4}\le x\le\frac{-21}{7}\)

\(\Rightarrow-7\le x\le-3\)

\(\Rightarrow x=\left\{-7;-6;-5;-4;-3\right\}\)

2)

a)\(\frac{108}{12}\le x\le\frac{91}{7}\)

\(\Rightarrow9\le x\le13\)

\(\Rightarrow x=\left\{9;10;11;12;13\right\}\)

\(a,\frac{4^5.2^{16}}{16^6}=\frac{2^{10}.2^{16}}{2^{24}}=\frac{2^{26}}{2^{24}}=2^2=4\)

\(b,\left(1-\frac{2}{5}\right)^2+|\frac{-3}{5}|+\frac{-7}{10}=\frac{9}{25}+\frac{3}{5}+\frac{-7}{10}=\frac{24}{25}-\frac{7}{10}=\frac{13}{50}\)

c, tt

a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)

   =1

b)=(1.2.3....8).(9-1-8)

   =(1.2.3....8).0

   =0

mik chỉ giải được zậy thôi.

t mik nha.

\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)

\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+3^{10}.2^{20}}\)

\(B=\frac{2^{19}.3^9+3^9.5.2^{18}}{2^{19}.3^9+3^{10}.2^{20}}\)

\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+3.2\right)}\)

\(B=\frac{7}{2.7}\)

\(B=\frac{1}{2}\)

\(C=\frac{2^{13}.4^{11}-16^9}{\left(3.2^{17}\right)^2}\)

\(C=\frac{2^{13}.2^{22}-2^{36}}{3^2.2^{34}}\)

\(C=\frac{2^{35}-2^{36}}{3^2.2^{34}}\)

\(C=\frac{2^{35}\left(1-2\right)}{3^2.2^{34}}\)

\(C=\frac{-2}{9}\)

\(D=\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)

\(D=\frac{2^{14}.2^8}{3.2^{15}.2^8-5.2^2.2^{20}}\)

\(D=\frac{2^{14}.2^8}{3.2^{23}-5.2^{22}}\)

\(D=\frac{2^{22}}{2^{22}\left(3.2-5\right)}\)

\(D=1\)

ai nhanh nhất mà trả lời dúng mik tặng 3 k

ta có: 4^7.2^8 / 3.2^15.16^2 - 5.2^2.(2^10)^2

      = 2^14.2^8 / 3.2^15.2^8 - 5.2^2.2^20

      = 2^22 / 6.2^22 - 5.2^22

      = 2^22 / 2^22 ( 6 - 5 )

      = 1

Câu 1 dễ mà :

1.2.3...9 - 1.2.3...8 - 1.2.3...7.8²

= 1.2.3...8.9 - 1.2.3...8.1 - 1.2.3...7.8.8

= 1.2.3...8.( 9 - 1 - 8 )

= 1.2.3...8.0

= 0

còn câu 2 và 3 thì sao

K ghi lại đề câu 2 nha :

\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)

\(=\frac{3^2.2^{36}}{11.2^{35}-2^{35}.2}\)

\(=\frac{3^2.2^{36}}{2^{35}\left(11-2\right)}\)

\(=\frac{3^2.2^{36}}{2^{35}.3^2}\)

\(=\frac{3^2}{3^2}.\frac{2^{36}}{2^{35}}\)

\(=1.2=2\)