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

\(\frac{4^6.3^4.9^5}{6^{12}}=\frac{\left(2^2\right)^6.3^4.\left(3^2\right)^5}{2^{12}.3^{12}}=\frac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\frac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)

\(\frac{4^6\cdot3^4\cdot9^5}{6^{12}}\)

\(=\frac{2^{12}\cdot3^4\cdot3^{10}}{2^{12}\cdot3^{12}}\)

\(=\frac{1\cdot3^{14}}{1\cdot3^{12}}\)

\(=\frac{3^{14}}{3^{12}}\)

\(=3^2\)

\(=9\)

f) \(\left(1:\frac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\frac{1}{49}\)

\(=\left(1\cdot7\right)^2:\left(2^6:2^5\right)\cdot\frac{1}{49}=7^2\cdot\frac{1}{2}\cdot\frac{1}{49}=49\cdot\frac{1}{49}\cdot\frac{1}{2}=\frac{1}{2}\)

g) \(\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{\left(2^2\right)^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot\left(3^2\right)^3+\left(2^3\right)^4\cdot3^5}\)

\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\left(3^5-3^6\right)}{2^{12}\left(3^6+3^5\right)}=\frac{2^{12}\left[3^5\left(1-3\right)\right]}{2^{12}\left[3^5\left(3+1\right)\right]}=\frac{2^{12}\cdot3^5\cdot\left(-2\right)}{2^{12}\cdot3^5\cdot4}=\frac{-2}{4}=-\frac{1}{2}\)

                                                               Bài giải

\(f,\text{ }\left(1\text{ : }\frac{1}{7}\right)^2\left[\left(2^2\right)^3\text{ : }2^5\right]\cdot\frac{1}{49}\)

\(=7^2\left(2^6\text{ : }2^5\right)\cdot\frac{1}{7^2}\)

\(=2\)

\(g,\text{ }\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^5\cdot\left(1-3\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}=-\frac{2}{4}=-\frac{1}{2}\)

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

\(a=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-\left(3^2\right)^{15}}{\left(2^2.3^{28}\right)}=\frac{11.3^{29}-3.^{30}}{2^2.328}\)

\(=\frac{3^{28}\left(11.3-3^2\right)}{2^2.3^{28}}=\frac{33-9}{4}=6\)

A=\(\frac{72^3.54^2}{108^4}=\frac{\left(2^3.3^2\right)^3.\left(2.3^3\right)^2}{\left(2^2.3^3\right)^4}=\frac{2^9.3^6.2^2.3^6}{2^8.3^{12}}=\frac{2^{11}.3^{12}}{2^8.3^{12}}=2^3=8\)

B= \(\frac{4^6.3^4.9^5}{6^{12}}=\frac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\frac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)

c) \(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)

1.

\(\frac{72^3\times54^2}{108^4}=\frac{\left(8\times9\right)^3\times\left(27\times2\right)^2}{\left(27\times4\right)^4}=\frac{\left(2^3\times3^2\right)^3\times\left(3^3\times2\right)^2}{\left(3^3\times2^2\right)^4}=\frac{\left(2^3\right)^3\times\left(3^2\right)^3\times\left(3^3\right)^2\times2^2}{\left(3^3\right)^4\times\left(2^2\right)^4}=\frac{2^9\times3^6\times3^6\times2^2}{3^{12}\times2^8}=2^3=8\)

2.

\(\frac{4^6\times3^4\times9^5}{6^{12}}=\frac{\left(2^2\right)^6\times3^4\times\left(3^2\right)^5}{\left(2\times3\right)^{12}}=\frac{2^{12}\times3^4\times3^{10}}{2^{12}\times3^{12}}=3^2=9\)

3.

\(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\times\left(2^8+1\right)}{2^2\times\left(2^8+1\right)}=2^3=8\)

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

\(A=\frac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)

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

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

\(A=\frac{3^{29}.8}{4.3^{28}}=3.2=6\)

\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)