\(\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right).\sqrt{2}\\ =\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|-\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)

\(\dfrac{2\sqrt{3}-3\sqrt{2}}{\sqrt{5-2\sqrt{6}}}=\dfrac{\sqrt{3}.\sqrt{2}.\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}}\\ =\dfrac{\sqrt{6}.\left(\sqrt{2}-\sqrt{3}\right)}{\left|\sqrt{2}-\sqrt{3}\right|}\\ =\dfrac{-\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=-\sqrt{6}\)

Kẻ đường cao \(AH\).

\(cot60^o=\dfrac{BH}{AH},cot50^o=\dfrac{CH}{AH}\)

\(cot60^o+cot50^o=\dfrac{BH+CH}{AH}=\dfrac{20}{AH}\)

\(\Leftrightarrow AH=\dfrac{20}{cot60^o+cot50^o}\)

\(S_{ABC}=\dfrac{1}{2}.20.\dfrac{20}{cot60^o+cot50^o}=\dfrac{200}{cot60^o+cot50^o}\)

\(\dfrac{a}{2-a}+\dfrac{b}{2-b}=1\Leftrightarrow a\left(2-b\right)+b\left(2-a\right)=\left(2-a\right)\left(2-b\right)\)

<=> 4a + 4b = 3ab + 4

Khi đó \(M=\sqrt{\left(4a+4b\right)^2-24ab-16}-3ab\)

\(=\sqrt{\left(3ab+4\right)^2-24ab-16}-3ab=\sqrt{\left(3ab\right)^2}-3ab=0\left(\text{vì }a;b>0\right)\)