\(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2+\sqrt{2}+\sqrt{2+\sqrt{3}}}\cdot\sqrt{2-\sqrt{2+\sqrt{3}}}\)

\(=\sqrt{\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{2+\sqrt{3}}\right)\cdot\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\cdot\left(2-\sqrt{2+\sqrt{3}}\right)}\)

\(=\sqrt{\left(2+\sqrt{3}\right)\cdot\left(4-\left(2+\sqrt{3}\right)\right)\cdot\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)

\(=\sqrt{\left(2+\sqrt{3}\right)\cdot\left(4-2-\sqrt{3}\right)\cdot\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)

\(=\sqrt{\left(2+\sqrt{3}\right)\cdot\left(2-\sqrt{3}\right)\cdot\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)

\(=\sqrt{\left(4-3\right)\cdot\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)

\(=\sqrt{1\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)

\(=\sqrt{2+\sqrt{2}+\sqrt{2+\sqrt{3}}}\)

\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)

\(=\dfrac{\dfrac{291}{385}}{\dfrac{388}{385}}\)

\(=\dfrac{291}{388}\)

\(=\dfrac{3}{4}\)

d) \(D=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}\)

\(=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}\)

\(=\dfrac{\left(2^{10}+1\right)\cdot2^{20}}{\left(1+2^{10}\right)\cdot2^{12}}\)

\(=\dfrac{\left(1024+1\right)\cdot2^8}{1+1024}\)

\(=\dfrac{1025\cdot2^8}{1025}\)

\(=2^8\)

\(=256\)

e) \(E=\dfrac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)

\(=\dfrac{5\cdot3\cdot2\cdot2^{14}\cdot3^{29}-5\cdot14^5\cdot4096}{9\cdot6\cdot6^{14}\cdot9^7-3\cdot2^2\cdot2^{15}\cdot7^5}\)

\(=\dfrac{5\cdot3^{30}\cdot2^{15}-20480\cdot14^5}{9^8\cdot6^{15}-3\cdot2^{17}\cdot7^5}\)

\(=\dfrac{5\cdot3^{30}\cdot32768-20480\cdot14^5}{9^8\cdot6^{15}-3\cdot2^{17}\cdot16807}\)

\(=\dfrac{163840\cdot3^{30}-20480\cdot14^5}{9^8\cdot6^{15}-50421\cdot2^{17}}\)

f) \(F=\dfrac{16^3\cdot3^{16}+120\cdot6^9}{4^6\cdot3^{12}+6^{11}}\)

\(=\dfrac{2^{12}\cdot3^{16}+20\cdot6\cdot6^9}{2^{12}\cdot3^{12}+6^{11}}\)

\(=\dfrac{3^4\cdot6^{12}+20\cdot6^{10}}{\left(2\cdot3\right)^{12}+6^{11}}\)

\(=\dfrac{81\cdot6^{12}+20\cdot6^{10}}{6^{12}+6^{11}}\)

\(=\dfrac{\left(81\cdot6^2+20\right)\cdot6^{10}}{\left(6+1\right)\cdot6^{11}}\)

\(=\dfrac{81\cdot36+20}{7\cdot6}\)

\(=\dfrac{2916+20}{7\cdot6}\)

\(=\dfrac{2936}{7\cdot6}\)

\(=\dfrac{1468}{7\cdot3}\)

\(=\dfrac{1468}{21}\)

a) \(A=\dfrac{12^3\cdot54^2}{108^4}\)

\(=\dfrac{12^3\cdot2^2\cdot\left(3^3\right)^2}{9^4\cdot12^4}\)

\(=\dfrac{4\cdot3^6}{\left(3^2\right)^4\cdot12}\)

\(=\dfrac{1\cdot3^5}{3^8}\)

\(=\dfrac{1}{3^3}\)

\(=\dfrac{1}{27}\)

b) \(B=\dfrac{3^{12}\cdot13+3^{12}\cdot3}{3^{11}\cdot2^4}\)

\(=\dfrac{13\cdot3^{12}+3\cdot3^{12}}{3^{11}\cdot2^4}\)

\(=\dfrac{16\cdot3^{12}}{3^{11}\cdot2^4}\)

\(=\dfrac{2^4\cdot3}{2^4}\)

\(=3\)

c) \(C=\dfrac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}\)

\(=\dfrac{13\cdot2^{10}+65\cdot2^{10}}{2^8\cdot104}\)

\(=\dfrac{78\cdot2^{10}}{2^8\cdot104}\)

\(=\dfrac{3\cdot2^2}{4}\)

\(=\dfrac{3\cdot4}{4}\)

\(=3\)

cái dấu ở giữa là phần hay ngăn ra 2 câu vậy

ko phải người thường thì là Yêu Tinh Chó

a) \(3^{-2}\cdot3^4\cdot3^n=3^7\) (1)

\(\Leftrightarrow3^{n+2}=3^7\)

\(\Leftrightarrow n+2=7\)

\(\Leftrightarrow n=7-2\)

\(\Leftrightarrow n=5\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{5\right\}\)

b) \(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\) (2)

\(\Leftrightarrow\left(2^{-1}+4\right)\cdot2^n=9\cdot2^5\)

\(\Leftrightarrow\left(\dfrac{1}{2}+4\right)\cdot2^n=9\cdot2^5\)

\(\Leftrightarrow\dfrac{9}{2}\cdot2^n=9\cdot2^5\)

\(\Leftrightarrow2^n=2^6\)

\(\Leftrightarrow n=6\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{6\right\}\)

Where did you go on your last vacation?