Gợi ý 

bn thực hiện phép tính tử mẫu bình thường , khi ra nhưng số trùng nhau bn gạch ra nháp cho đến nhưng số tối giản là ra nha 

chúc bn

học tốt

A = \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

  = \(\frac{3^{10}\left(11+5\right)}{3^9.2^4}\)

 = \(\frac{3^{10}.16}{3^9.2^4}\)

 = \(\frac{3^{10}.2^4}{3^9.2^4}=3\)

B = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)

  = \(\frac{2^{10}\left(13+65\right)}{2^8.104}\)

  = \(\frac{2^{10}.78}{2^8.104}\)

  = \(\frac{2^{10}.13.2.3}{2^8.2^3.13}\)

 = \(\frac{2^{11}.13.3}{2^{11}.13}=3\)

C = \(\frac{72^3.54^2}{108^4}\)

  = \(\frac{\left(2^3.3^2\right)^3.\left(2.3^3\right)^2}{\left(2^2.3^3\right)^4}\)

  = \(\frac{2^9.3^6.2^2.3^6}{2^8.3^{12}}\)

 = \(\frac{2^{11}.3^{12}}{2^8.3^{12}}\)

= \(2^3=8\)

:):):)lop1

a, A = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)

\(A=\frac{2^{10}\left(13+65\right)}{2^8.2^2.26}=\frac{2^{10}.78}{2^{10}.26}=\frac{78}{26}=3\)

Vậy A = 3

b, \(B=\frac{72^3.54^2}{108^4}=\frac{72^3.54^2}{\left(54.2\right)^4}=\frac{72^3.54^2}{54^4.2^4}=\frac{72^3}{54^2.2^4}=\frac{\left(8.9\right)^3}{\left(6.9\right)^2.2^4}\)

\(=\frac{\left(2^3\right)^3.9^3}{6^2.9^2.2^4}=\frac{2^9.9^3}{2^2.3^2.9^2.2^4}=\frac{2^9.9^3}{2^6.9^3}=\frac{2^9}{2^6}=2^3=8\)

Vậy B = 8

c, \(C=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}.3^{30}}{2^2.3^{28}}=\frac{11.3^{29}.3.3^{29}}{2^2.3^{28}}=\frac{\left(11-3\right)3^{29}}{2^2.3^{28}}\)

\(=\frac{2^3.3^{29}}{2^2.3^{28}}=2.3=6\)

Vậy C = 6

d, \(D=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(3.2^{18}\right)^2}{11.2^{35}-\left(2^4\right)^9}=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}=\frac{3^2.2^{36}}{\left(11-2\right)2^{35}}=\frac{3^2.2}{9}=2\)

Vậy D = 2

\(A=\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{3^{10}\cdot\left(11+5\right)}{3^9\cdot16}=\dfrac{3^{10}\cdot16}{3^9\cdot16}=3\)

\(B=\dfrac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\dfrac{2^{10}\cdot\left(13+65\right)}{2^8\cdot2^2\cdot26}=\dfrac{2^{10}\cdot78}{2^{10}\cdot26}=3\)

\(C=\dfrac{72^3\cdot54^2}{108^4}=\dfrac{\left(2^3\cdot3^2\right)^3\cdot\left(2\cdot3^3\right)^2}{\left(3^3\cdot2^2\right)^4}\\ =\dfrac{2^9\cdot3^6\cdot2^4\cdot3^6}{3^{12}\cdot2^8}=\dfrac{2^{13}\cdot3^{12}}{3^{12}\cdot2^8}=2^5=32\)

\(D=\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\dfrac{11\cdot3^{29}-\left(3^2\right)^{15}}{2^2\cdot3^{28}}=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}\\ =\dfrac{3^{29}\cdot\left(11-3\right)}{2^2\cdot3^{28}}=\dfrac{3^{29}\cdot8}{4\cdot3^{28}}=3\cdot2=6\)

mấy bn lm giúp mk nha , mk tk cho 3 tk 

Bài 35 :

\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}\)

\(A=\frac{2^{10}.\left(13+65\right)}{2^8.104}\)

\(A=\frac{2^8.2^2.98}{2^8.104}\)

\(A=\frac{2^8.4.98}{2^8.4.26}\)

\(A=\frac{49}{13}\)

Vậy \(A=\frac{49}{13}\)

\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)

\(B=\frac{11.3^{29}-9^{15}}{2^2.\left(3^{14}\right)^2}\)

\(B=\frac{11.3^{29}-9^{15}}{2^2.3^{28}}\)

\(B=\frac{11.3^{29}-\left(3^2\right)^{15}}{4.3^{28}}\)

\(B=\frac{11.3^{29}-3^{30}}{4.3^{28}}\)

\(B=\frac{11.3^{29}-3^{29}.3}{4.3^{28}}\)

\(B=\frac{3^{29}.\left(11-3\right)}{4.3^{28}}\)

\(B=\frac{3^{29}.8}{4.3^{28}}\)

\(B=\frac{3^{28}.3.4.2}{4.3^{28}}\)

\(B=3.2\)

\(B=6\)

Vậy B = 6 

A = 2^10 . 13 + 2^10 . 65 / 2^8 . 104

   = 2^10 ( 13 + 65 ) / 2^8 . 104 = 2^10 . 78 / 2^8 . 104 = 2^8 . 2^2 . 78 / 2^8 . 104 = 2^8 . 4 . 78 / 2^8 . 104 = 2^8 . 312 / 2^8 . 104

   = 312/104

   = 3

B = 11 . 3^22 . 3^7 - 9^15 / ( 2.3^14)^2

   = 11 . 3^29 - (3^2)^15 / ( 3.2^14)^2

   = 11 . 3^29 - 3^30 / ( 3. 2 )^28

    = ( 8 + 3 ) . 3^29 - 3^30 / ( 3. 2)^28

    = 8 . 3^29 + 3.3^29 - 3^30 / ( 3.2)^28

     = 8 . 3^29 + 3^30 - 3^30 / ( 3 . 2)^28

    = 8 . 3^29 / 3^28 . 2^28

     = 2^3 . 3 / 2^28

     = 3/ 2^25

2 bạn ai làm đúng vậy

\(A=\frac{72^2.54^2}{1084}=\frac{\left(72.54\right)^2}{1084}=\frac{3888^2}{1084}=\frac{3888.3888}{1084}=\frac{972.3888}{271}=\frac{3779136}{271}\)

\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.16}{3^9.16}=3\)

\(B=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.78}{2^8.104}=\frac{2^{10}.26.3}{2^8.2^2.26}=3\)

\(C=\frac{72^3.52^4}{108^4}=\frac{\left(3^2.2^3\right)^3.\left(13.2^2\right)^4}{\left(3^3.2^2\right)^4}=\frac{3^6.2^9.13^4.2^8}{3^{12}.2^8}=\frac{2^9.13^4}{3^6}\)

\(D=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}\left(33-1\right)}{2^2.3^{28}}=\frac{3^{29}.2^5}{2^2.3^{28}}=3.8=24\)