\(x^3=6+3x\sqrt[3]{\left(3+\sqrt{\frac{368}{27}}\right)\left(3-\sqrt{\frac{368}{27}}\right)}\Leftrightarrow x^3=6+3x.\sqrt[3]{9-\frac{368}{27}}\Leftrightarrow x^3+5x-6=0\)

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

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

b: \(=3\sqrt{3}-\dfrac{6}{\sqrt{3}}+1-\sqrt{3}\)

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

\(\dfrac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)

b. 

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

\(=\sqrt{3-\sqrt{5}}.\sqrt{9-5}+\sqrt{3+\sqrt{5}}.\sqrt{9-5}\)

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

\(=\sqrt{\left(\sqrt{10}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}\)

\(=\sqrt{10}-\sqrt{2}+\sqrt{10}+\sqrt{2}=2\sqrt{10}\)

c.

\(\dfrac{a-\sqrt{b}}{\sqrt{b}}:\dfrac{\sqrt{b}}{a+\sqrt{b}}=\dfrac{\left(a-\sqrt{b}\right)\left(a+\sqrt{b}\right)}{\sqrt{b}.\sqrt{b}}=\dfrac{a^2-b}{b}\)

\(1,=2\sqrt{3}-3\sqrt{3}+4\sqrt{3}=3\sqrt{3}\\ 2,=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right):5=\dfrac{2\sqrt{6}}{5}-\dfrac{2\sqrt{5}}{5}\\ 3,=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-\dfrac{9\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\\ 4,Sửa:\dfrac{1}{\sqrt{5}-\sqrt{3}}-\dfrac{1}{\sqrt{5}+\sqrt{3}}\\ =\dfrac{\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)

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

2) \(=\left(2\sqrt{6}+2\sqrt{5}-4\sqrt{5}\right)=\dfrac{2\sqrt{6}}{5}+\dfrac{2\sqrt{5}}{5}-\dfrac{4\sqrt{5}}{5}\)

3) \(=6\sqrt{3}-\dfrac{4\sqrt{3}}{3}-4\sqrt{3}-\dfrac{5\sqrt{3}}{3}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)

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

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

a: \(=\dfrac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)

b: \(=\dfrac{\sqrt{10}\left(\sqrt{11}+\sqrt{7}\right)}{\sqrt{2}\left(\sqrt{11}+\sqrt{7}\right)}=\sqrt{\dfrac{10}{2}}=\sqrt{5}\)

c: \(=\dfrac{\sqrt{6}\left(\sqrt{7}-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{7}-\sqrt{6}\right)}=\sqrt{\dfrac{6}{3}}=\sqrt{2}\)

1) \(\dfrac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{9\sqrt{5}+3\sqrt{9\cdot3}}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{9\sqrt{5}+3\cdot3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{9\cdot\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{9}{1}=9\)

2) \(\dfrac{\sqrt{110}+\sqrt{70}}{\sqrt{22}+\sqrt{14}}\)

\(=\dfrac{\sqrt{10}\cdot\sqrt{11}+\sqrt{10}\cdot\sqrt{7}}{\sqrt{2}\cdot\sqrt{11}+\sqrt{2}\cdot\sqrt{7}}\)

\(=\dfrac{\sqrt{10}\cdot\left(\sqrt{11}+\sqrt{7}\right)}{\sqrt{2}\cdot\left(\sqrt{11}+\sqrt{7}\right)}\)

\(=\dfrac{\sqrt{10}}{\sqrt{2}}=\sqrt{\dfrac{10}{2}}\)

\(=\sqrt{5}\)

3) \(\dfrac{\sqrt{42}-6}{\sqrt{21}-\sqrt{18}}\)

\(=\dfrac{\sqrt{6}\cdot\sqrt{7}-\sqrt{6}\cdot\sqrt{6}}{\sqrt{3}\cdot\sqrt{7}-\sqrt{3}\cdot\sqrt{6}}\)

\(=\dfrac{\sqrt{6}\cdot\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{3}\cdot\left(\sqrt{7}-\sqrt{3}\right)}\)

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

\(=\sqrt{2}\)

a) \(5\sqrt{27}+3\sqrt{48}-2\sqrt{12}-6\sqrt{3}\)

\(=15\sqrt{3}+12\sqrt{3}-4\sqrt{3}-6\sqrt{3}\)

\(=17\sqrt{3}\)

b) \(\dfrac{2}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+6\sqrt{3}\)

\(=6-3\sqrt{3}+4+\sqrt{3}+6\sqrt{3}\)

\(=4\sqrt{3}+10\)