Lời giải:
a.

$=2\sqrt{5}-9\sqrt{5}-2\sqrt{5}=(2-9-2)\sqrt{5}=-9\sqrt{5}$

b.

$=36\sqrt{6}-2\sqrt{6}+6\sqrt{6}=(36-2+6)\sqrt{6}=40\sqrt{6}$

`a)\root[3]{135}/\root[3]{5}-\root[3]{54}.\root[3]{4}`

`=\root[3]{135/5}-\root[3]{54.4}`

`=\root[3]{27}-\root[3]{216}`

`=3-6=-3`

`b)(\root[3]{25}-\root[3]{10}+\root[3]{4})(\root[3]{5}+\root[3]{2})`

`=5+\root[3]{50}-\root[3]{50}-\root[3]{20}+\root[3]{20}+2`

`=7`.

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

                             \(=\dfrac{\sqrt{3}-\sqrt{6}+\sqrt{6}-\sqrt{12}}{1-2}=\sqrt{12}-\sqrt{3}\)

b) \(Q=\left(\sqrt{75}-\dfrac{3}{2}:\sqrt{3}-\sqrt{48}\right)\cdot\sqrt{\dfrac{16}{3}}\)

        \(=\left(5\sqrt{3}-\dfrac{3}{2}\cdot\dfrac{1}{\sqrt{3}}-4\sqrt{3}\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\sqrt{3}\left(5-\dfrac{1}{2}-4\right)\cdot\dfrac{4}{\sqrt{3}}\)

        \(=\left(1-\dfrac{1}{2}\right)\cdot4=2\)

Lời giải:

\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{16}-2}-\frac{12}{3-\sqrt{16}}\right).(\sqrt{6}+11)=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4}{4-2}-\frac{12}{3-4}\right)(\sqrt{6}+11)\)

\(=\left(\frac{15(\sqrt{6}-1)}{6-1}+2+12\right)(\sqrt{6}+11)=(3\sqrt{6}-3+14)(\sqrt{6}+11)\)

\(=(3\sqrt{6}+11)(\sqrt{6}+11)\)

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\)\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+2+2}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)

a: \(=4\sqrt[3]{2}-9\sqrt[3]{2}++6\sqrt[3]{2}=\sqrt[3]{2}\)

b: \(=6\sqrt[3]{3}-15\sqrt[3]{3}+16\sqrt[3]{3}=7\sqrt[3]{3}\)

c: \(=-7\sqrt[3]{3}+3\sqrt[3]{3}+6\sqrt[3]{3}=2\sqrt[3]{3}\)

d: \(=8\sqrt[3]{5}-10\sqrt[3]{5}+2=-2\sqrt[3]{5}+2\)

A: \(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\)

\(=3-3\cdot\dfrac{5\sqrt{2}}{3}+6\sqrt{2}-3\)

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

b: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\cdot\sqrt{\dfrac{16}{3}}\)

\(=\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4}{\sqrt{3}}\)

\(=2-\sqrt{3}+\sqrt{3}+1-4\sqrt{3}\)

\(=3-4\sqrt{3}\)

\(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\\ =3-3\cdot\dfrac{1}{3}\sqrt{25\cdot2}+3\sqrt{4\cdot2}-3\\ =3-1\cdot5\sqrt{2}+3\cdot2\sqrt{2}-3\\ =3-5\sqrt{2}+6\sqrt{2}-3\\ =\sqrt{2}\)

\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\sqrt{\dfrac{16}{3}}\\ =\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4\sqrt{3}}{3}\\ =2-\sqrt{3}+\sqrt{3}+1-8\sqrt{3}\\ =3-8\sqrt{3}\)

\(A=\dfrac{\sqrt[4]{7\sqrt[3]{54}+15\sqrt[3]{128}}}{\sqrt[3]{\sqrt[4]{32}}+\sqrt[3]{9\sqrt[4]{162}}}\)

\(\Leftrightarrow A=\dfrac{\sqrt[4]{7\sqrt[3]{3^3.2}+15\sqrt[3]{4^3.2}}}{\sqrt[3]{\sqrt[4]{2^4.2}}+\sqrt[3]{9\sqrt[4]{3^4.2}}}\)

\(\Leftrightarrow A=\dfrac{\sqrt[4]{7.3\sqrt[3]{2}+15.4\sqrt[3]{2}}}{\sqrt[3]{2\sqrt[4]{2}}+\sqrt[3]{9.3\sqrt[4]{2}}}\)

\(\Leftrightarrow A=\dfrac{\sqrt[4]{21\sqrt[3]{2}+60\sqrt[3]{2}}}{\sqrt[3]{2\sqrt[4]{2}}+\sqrt[3]{3^3\sqrt[4]{2}}}\)

\(\Leftrightarrow A=\dfrac{\sqrt[4]{81\sqrt[3]{2}}}{\sqrt[3]{\sqrt[4]{2}}\left(\sqrt[3]{2}+3\right)}=\dfrac{3\sqrt[4]{\sqrt[3]{2}}}{\sqrt[3]{\sqrt[4]{2}}\left(\sqrt[3]{2}+3\right)}\)

\(\Leftrightarrow A=\dfrac{3}{\sqrt[3]{2}+3}\)