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

\(A=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)

\(A=\dfrac{11.3^{29}-3^{29}.3}{2^2.3^{28}}\)

\(A=\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\)

\(A=\dfrac{3^{29}.8}{2^2.3^{28}}\)

\(A=\dfrac{3.8}{4}=6\)

vậy \(A=6\)

b) \(B=\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

\(B=\dfrac{3^2.\left(2^2\right)^2.\left(2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

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

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

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

\(B=\dfrac{3^2.2^{36}}{2^{35}.9}\)

\(B=\dfrac{3^2.2}{9}\)

\(B=\dfrac{9.2}{9}\)

\(B=2\)

vậy \(B=2\)

\(\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

\(=\dfrac{\left(3.2^2.2^{16}\right)^2}{11.2^2.\left(2^2\right).11-\left(2^4\right)^9}\)

\(=\dfrac{3^2.\left(2^2\right)^2.\left(2^{16}\right)^2}{11.2^2.2^{22}-2^{36}}\)

\(=\dfrac{3^2.2^4.2^{32}}{11.2^{24}-2^{36}}\)

\(=\dfrac{3^2.2^{34}}{11.2^{24}-2^{36}}\)

\(=\dfrac{3^2.2^{24}.2^{10}}{11.2^{24}-2^{12}.2^{24}}\)

\(=\dfrac{3^2.2^{24}.2^{10}}{\left(11-2^{12}\right).2^{24}}\)

\(=\dfrac{3^2.2^{10}}{11-2^{12}}\)

\(=\dfrac{3^2\cdot2^{36}}{11\cdot2^{13}\cdot2^{22}-2^{36}}=\dfrac{3^2\cdot2^{36}}{2^{35}\cdot9}=2\)

\(A=\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)

\(B=\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=2\)

\(C=\frac{4^5\cdot9^{4-2\cdot6^9}}{2^{10}\cdot3^8+6^8\cdot20}=0\)

A=\(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)

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a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)

\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)

b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)

\(=\left[3-4\right]^2=1\)

c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)

\(=121-48=73\)

d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)

\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)

\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)

\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)

e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)

\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)

\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)

Ta có: \(A=\dfrac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{5\cdot2^{18}\cdot3^{18}\cdot2^{12}-2\cdot2^{28}\cdot3^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}-7\cdot2^{28}\cdot3^{18}\cdot2}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2\cdot2^{28}\cdot3^{18}}{2^{28}\cdot3^{18}\cdot\left(5-7\cdot2\right)}\)

\(=\dfrac{2^{28}\cdot3^{18}\cdot\left(5\cdot2^2-2\right)}{2^{28}\cdot3^{18}\cdot\left(5-14\right)}\)

\(=\dfrac{20-2}{-9}=\dfrac{18}{-9}=-2\)

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

\(\dfrac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=\dfrac{\left(3\cdot2^2\cdot2^{16}\right)^2}{11\cdot2^2\cdot\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\dfrac{3^2\cdot\left(2^2\right)^2\cdot\left(2^{16}\right)^2}{11\cdot2^2\cdot2^{22}-2^{36}}=\dfrac{3^2\cdot2^4\cdot2^{32}}{11\cdot2^{24}-2^{36}}\)

\(=\dfrac{3^2\cdot2^{34}}{11\cdot2^{24}-2^{36}}=\dfrac{3^2\cdot2^{24}\cdot2^{10}}{11\cdot2^{24}-2^{12}\cdot2^{24}}\)

\(=\dfrac{3^2\cdot2^{24}\cdot2^{10}}{\left(11-2^{12}\right)\cdot2^{24}}=\dfrac{3^2\cdot2^{10}}{11-2^{12}}\)