Ta có: \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\\ =2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)

\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)

\(=-13\sqrt{3}\)

1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)

\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)

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

\(=5\sqrt{3}\)

2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)

\(=\dfrac{2\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}\)

\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)

\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)

\(=\dfrac{-\sqrt{3}-5}{11}\)

3) Ta có: \(\sqrt{\dfrac{2}{5}}\)

\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)

\(=\dfrac{\sqrt{10}}{5}\)

a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

\(=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)

b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)

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

\(=1-5-2\sqrt{6}\)

\(=-4-2\sqrt{6}\)

\(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}=\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}=\left|3-\sqrt{6}\right|+\left|\sqrt{24}-3\right|=3-\sqrt{6}+\sqrt{24}-3=2\sqrt{6}-\sqrt{6}=\sqrt{6}\)

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}=-\dfrac{\sqrt{2}\left(\sqrt{6}-4\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\dfrac{-\sqrt{2}}{\sqrt{3}}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=-\dfrac{\sqrt{6}}{2}\).

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

^{Bạn tham khảo ctrl ạ!}

1.\(\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}=3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)

2.\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-2\sqrt{x}+1}{2}\)

\(P=\left(\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)

\(P=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)

\(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

\(a,\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}\\ =3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}\\ =-\sqrt{3}\)

\(b,P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-2\sqrt{x}+1}{2}\\ =\dfrac{\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\\ =\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\\ =\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

\(=\sqrt{3}\left(2\sqrt{3}+3\sqrt{3}-\sqrt{3}\right)=\sqrt{3}\cdot4\sqrt{3}=12\)

\(=\left(\sqrt{3}\cdot\sqrt{4}+\sqrt{9}\cdot\sqrt{3}-\sqrt{3}\right)\cdot\sqrt{3}\\ =3\cdot\left(2+3-1\right)=12\)

a/\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)

b/ \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)

\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)

\(=2\sqrt{3}\cdot6\sqrt{3}=2\cdot6\cdot3=36\)

c/ \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)

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

\(=1+2\sqrt{3}+3-2\)

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

d/ \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{8-4\sqrt{10}+5}-\sqrt{45+12\sqrt{10}+8}\)

\(=\sqrt{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2\cdot5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\cdot2\sqrt{5\cdot2}+\left(2\sqrt{2}\right)^2}\)

\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)

\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}\)

\(=-4\sqrt{5}\)

#)Giải :

 \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)