\(\frac{72}{55}\)

\(A=\frac{\frac{3}{2}+\frac{2}{5}+\frac{1}{10}}{\frac{3}{2}+\frac{2}{3}+\frac{1}{12}}\)

\(\Rightarrow A=\frac{\frac{15}{10}+\frac{4}{10}+\frac{1}{10}}{\frac{18}{12}+\frac{8}{12}+\frac{1}{12}}=\frac{\frac{20}{10}}{\frac{27}{12}}=\frac{2}{\frac{9}{4}}=2:\frac{9}{4}=2.\frac{4}{9}=\frac{8}{9}\)

! Ko bt có đúng ko nx  @@@

~ Học tốt 

# Chiyuki Fujito

\(A=\frac{\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right)}{\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)}\)

\(A=\frac{\left(\frac{15}{10}-\frac{4}{10}+\frac{1}{10}\right)}{\left(\frac{18}{12}-\frac{8}{12}+\frac{1}{12}\right)}\)

\(A=\frac{\frac{6}{5}}{\frac{11}{12}}=\frac{6}{5}:\frac{11}{12}=\frac{6}{5}\times\frac{12}{11}\)

\(A=\frac{72}{55}\)

1. \(A=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)=\frac{6}{5}:\frac{11}{12}=\frac{6}{5}.\frac{12}{11}=\frac{72}{55}\)

2. 2^x+2 . 3^x+1 . 5^x = 10800

=> 2^x . 2² . 3^x . 3 . 5^x = 10800

=> ( 2 . 3 . 5 )^x . 12 = 10800

=> 30^x = 900

=> 30^x = 30²

=> x = 2

\(A=\frac{10}{3}+\frac{10}{9}+\frac{10\sqrt{5}}{3\sqrt{36\sqrt{6}}}\)

\(A=\frac{40}{9}+\frac{10\sqrt{5}}{18\sqrt{\sqrt{6}}}\)

trục căn thứa là ra nha bạn

\(A=\frac{40}{9}+\frac{10\sqrt{5}}{18\sqrt{\sqrt{6}}}\)

Ta có:

\(A=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\cdot\sqrt{\frac{5}{12}-\frac{1}{16}}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\cdot\sqrt{\frac{17}{48}}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{51}}{12}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{17}}{12}\)

\(A=\frac{4\sqrt{3}+2\sqrt{2}+\sqrt{17}}{12}\)

Ta có: \(\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\sqrt{\frac{5}{12}-\frac{\sqrt{6}}{6}}=\sqrt{\frac{5-2\sqrt{6}}{12}}\)

Vì \(5-2\sqrt{6}=3-2\sqrt{3}.\sqrt{2}+2=\left(\sqrt{3}\right)^2-2\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2\)\(\Rightarrow5-2\sqrt{6}=\left(\sqrt{3}-\sqrt{2}\right)^2\)

Như vậy: \(\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{1}{2\sqrt{3}}\left(\sqrt{3}-\sqrt{2}\right)\)

Lại có: \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{\sqrt{3}}.\frac{1}{2\sqrt{3}}\left(\sqrt{3}-\sqrt{2}\right)\)

Rút gọn ta được \(A=\frac{\sqrt{3}}{2}\)

\(A=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}\sqrt{\frac{5}{12}-\frac{\sqrt{6}}{6}}\)\(=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}\sqrt{\frac{5-2\sqrt{6}}{12}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}\)

\(=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{3}}\left(do\sqrt{3}-\sqrt{2}>0\right)\)\(=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{1}{6}\left(\sqrt{3}-\sqrt{2}\right)=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}+\frac{\sqrt{3}}{3}-\frac{\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)

KQ:\(\frac{1}{5}\)

cho tớ xin cách lm