Bài giải

\(d,\text{ }\frac{72^3\cdot54^2}{108^4}=\frac{3^6\cdot2^9\cdot3^6\cdot2^2}{3^{12}\cdot2^8}=2^3=9\)

\(e,\text{ }\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\frac{3^{10}\left(11+5\right)}{3^9\cdot2^4}=\frac{3\cdot16}{2^4}=\frac{3\cdot2^4}{2^4}=3\)

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

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

\(=\frac{3^{10}.16}{3^9.2^4}\)

\(=\frac{3^{10}.2^4}{3^9.2^4}=3\)

3^10.(11+5)      =3.16

3^9 . 2^4            1.16                                                                                                                                                                                                     bỏ số 16 thì được kết quả bằng 3

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

=3 nhaaaaa

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

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

\(=\frac{3^{10}.16}{3^9.2^4}\)

\(=\frac{3^{10}.2^4}{3^9.2^4}\)

\(=3\)

a)\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{2^4}=\frac{3.2^4}{2^4}=3\)

b)\(B=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.2^3.13}=\frac{2^{10}.78}{2^{11}.13}=3\)

c)\(C=\frac{4^9.36+64^4}{16^4.100}=\frac{2^{18}.2^2.3^2+2^{24}}{2^{16}.2^2.5^2}=\frac{2^{20}\left(3^2+2^4\right)}{2^{18}.5^2}=\frac{2^2.25}{25}=4\)

\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3.16}{16}=3\)

\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{4.78}{104}=\frac{312}{104}=3\)

~ Chúc em học tốt ~

Ta có:

B = \(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.2^4}=\frac{3^{10}.16}{3^9.2^4}=\frac{3^{10}.2^4}{3^9.2^4}=3\)

C = \(\frac{2^{10}.39+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(39+65\right)}{2^8.104}=\frac{2^{10}.104}{2^8.104}=4\)

Ta thấy : 3 < 4 => B < C

A= 3^10.( 11+5 ) / 3^9. 2^4 

A= 3^10. 16 /3 ^9 . 16

A= 3^10/3^9= 3

B= 2^10. ( 13 +65) / 2^8.104

B= 2^10. 78/ 2^8.104

B= 2 ^10.2.39/ 2^8 .2.52

B= 2^11.39/ 2^9.52

B= 2 ^ 2. (39/ 52)

B= 4 . 39/52 = 3

C= (2^3.3^2)^3.( 2.3.3^2 )^2 / (2^2.3^3)^4

C= 2^9.3^6/ 2^2.3^2.3^4

C= 2^9.3^6/ 2^2.3^6

C= 2^9/2^2= 2^5=32

D= 11.3^29-3^30/ 2^2.3^28

D= 3^29.(1- 3)/ 2^2.3^28

D= 3^29.(-2)/ 2^2.3^28

D= 3. (-2/2^2)

D = 3. (-1/2)= -3/2

\(\frac{6:\frac{3}{5}-1\frac{1}{6}\times\frac{6}{7}}{4\frac{1}{5}\times\frac{10}{11}+5\frac{2}{11}}\)

\(=\frac{\frac{6}{1}:\frac{3}{5}-\frac{7}{6}\times\frac{6}{7}}{\frac{21}{5}\times\frac{10}{11}\times\frac{57}{11}}\)

\(=\frac{\frac{6}{1}\times\frac{5}{3}-1}{\frac{210}{55}+\frac{57}{11}}\)

\(=\frac{\frac{30}{3}-1}{\frac{42}{11}+\frac{57}{11}}\)

\(=\frac{10-1}{\frac{99}{11}}\)

\(=\frac{9}{9}\)

\(=1\)

\(6:\frac{3}{5}-1\frac{1}{6}\)X \(\frac{6}{7}\)                                                     \(4\frac{1}{5}\)X \(\frac{10}{11}+5\frac{2}{11}\)

\(=\frac{33}{5}-\frac{7}{6}\)X \(\frac{6}{7}\)                                             \(=\)      \(\frac{21}{5}\)X  \(\frac{10}{11}+\frac{57}{11}\) 

\(=\frac{33}{5}-1\)                                                              \(=\frac{42}{11}+\frac{57}{11}\)

\(=\frac{28}{5}\)                                                                       \(=\frac{99}{11}=9\)