\(=\dfrac{1}{\sqrt{3}}\cdot\dfrac{6}{3-\sqrt{3}}=\dfrac{6}{3\left(\sqrt{3}-1\right)}=\dfrac{2}{\sqrt{3}-1}=\sqrt{3}+1\)

\(\dfrac{\sqrt{3}}{3}.\dfrac{6}{3-\sqrt{3}}=\dfrac{6\sqrt{3}}{3\sqrt{3}\left(\sqrt{3}-1\right)}=\dfrac{2}{\sqrt{3}-1}\)

a: =2015+6-5=2016

b: =10căn 2+5căn 2-6căn 2=9căn 2

c: =3căn 3-4căn 3-5căn 3=-6căn 3

d: =2căn 3+3căn 3-4căn 3=căn 3

\(A=2015+6-5==2015+1=2016\)

\(B=5\sqrt{2^3}+\sqrt{5^2.2}-2\sqrt{3^2.2}\\ =10\sqrt{2}+5\sqrt{2}-6\sqrt{2}\\ =\left(10+5-6\right)\sqrt{2}=9\sqrt{2}\)

\(C=\sqrt{3^3}-2\sqrt{2^2.3}-\sqrt{5^2.3}\\ =3\sqrt{3}-4\sqrt{3}-5\sqrt{3}\\ =\left(3-4-5\right)\sqrt{3}=-6\sqrt{3}\)

\(D=\sqrt{2^2.3}+\sqrt{3^3}-\sqrt{4^2.3}\\ =2\sqrt{3}+3\sqrt{3}-4\sqrt{3}\\ =\left(2+3-4\right)\sqrt{3}=\sqrt{3}\)

\(B=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-\dfrac{36}{x-9}\)

\(=\dfrac{\left(\sqrt{x}+3\right)^2+\left(\sqrt{x}-3\right)^2-36}{x-9}\)

\(=\dfrac{x+6\sqrt{x}+9+x-6\sqrt{x}+9-36}{x-9}\)

\(=\dfrac{2x-18}{x-9}=\dfrac{2\left(x-9\right)}{x-9}=2\)

ĐKXĐ : \(x\ge0;x\ne9\)

Ta có : \(B=\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\dfrac{x+6\sqrt{x}+9+x-6\sqrt{x}+9-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=2\)

\(3\sqrt{144}-5\sqrt{49}+\dfrac{1}{2}\sqrt{36}\)

\(=3.12-5.7+\dfrac{1}{2}.6\)

\(=36-35+3=4\)

\(A=\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}\)

=>\(A^2=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{4}\)

=>A^2=2căn 7-4

=>A=2căn 7-4

=>\(M=\dfrac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)