\(E=\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)

\(=\sqrt{9-2.3\sqrt{3}+3}+\sqrt{12-2.3.2\sqrt{3}+9}\)

\(=\sqrt{\left(9-3\right)^2}+\sqrt{\left(12-9\right)^2}\)

\(=\sqrt{6^2}+\sqrt{3^2}\)

\(=6+3=9\)

e) Ta có: \(E=\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)

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

\(=\sqrt{3}\)

f) Ta có: \(F=\sqrt{21-12\sqrt{3}}+\sqrt{21+12\sqrt{3}}\)

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

\(=4\sqrt{3}\)

h) Ta có: \(H=\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)

\(=3+\sqrt{5}+3-\sqrt{5}\)

=6

\(F=\sqrt{21-12\sqrt{3}}+\sqrt{21+12\sqrt{3}}\)

\(=\sqrt{12-2.3.2\sqrt{3}+9}+\sqrt{12+2.3.2\sqrt{3}+9}\)

\(=\sqrt{\left(12-9\right)^2}+\sqrt{\left(12+9\right)^2}\)

\(=\sqrt{3^2}+\sqrt{21^2}\)

\(=3+21=24\)

\(e,=\dfrac{\left(3+\sqrt{2}\right)\left(2\sqrt{2}+1\right)}{7}-\sqrt{\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}\\ =\dfrac{7\sqrt{2}+7}{7}-\dfrac{\sqrt{2}+1}{1}=\sqrt{2}+1-\sqrt{2}-1=0\)

\(f,=\sqrt{\dfrac{\left(2\sqrt{3}-3\right)^2}{\left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right)}}\left(2+\sqrt{3}\right)\\ =\dfrac{\left(2\sqrt{3}-3\right)\left(2+\sqrt{3}\right)}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}=1\)

\(h,=\sqrt{\dfrac{\left(3\sqrt{5}-1\right)\left(2\sqrt{5}-3\right)}{20-9}}\left(\sqrt{2}+\sqrt{10}\right)\\ =\sqrt{\dfrac{2\left(33-11\sqrt{5}\right)}{11}}\left(\sqrt{5}+1\right)\\ =\sqrt{\dfrac{22\left(3-\sqrt{5}\right)}{11}}\left(\sqrt{5}+1\right)\\ =\sqrt{6-2\sqrt{5}}\left(\sqrt{5}+1\right)=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=4\)

c: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

c) \(\dfrac{4}{5}+\dfrac{4}{-18}=\dfrac{4}{5}+\dfrac{-4}{18}=\dfrac{72}{90}+\dfrac{-20}{90}=\dfrac{52}{45}=\dfrac{26}{45}\)

Giải:

c) \(\dfrac{4}{5}+\dfrac{4}{-18}=\dfrac{4}{5}+\dfrac{-2}{9}=\dfrac{36}{45}+\dfrac{-10}{45}=\dfrac{36+\left(-10\right)}{45}=\dfrac{26}{45}\) 

d) \(\dfrac{-18}{24}+\dfrac{15}{-21}=\dfrac{-3}{4}+\dfrac{-5}{7}=\dfrac{-21}{28}+\dfrac{-20}{28}=\dfrac{-21+\left(-20\right)}{28}=\dfrac{-41}{28}\) 

g) \(\dfrac{3}{5}-\dfrac{5}{6}=\dfrac{18}{30}-\dfrac{25}{30}=\dfrac{18-25}{30}=\dfrac{-7}{30}\) 

h) \(\dfrac{-5}{9}-\dfrac{-5}{12}=\dfrac{-20}{36}-\dfrac{-15}{36}=\dfrac{-20-\left(-15\right)}{36}=\dfrac{-5}{36}\)

a: Xét tứ giác OBAC có 

\(\widehat{OBA}+\widehat{OCA}=180^0\)

Do đó: OBAC là tứ giác nội tiếp

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