Bài 2:

a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))

\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)

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

\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\) (thỏa)

Vậy...

b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))

\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=2\) (tm đk)

Vậy...

c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))

\(\Leftrightarrow x-2\sqrt{x-1}=x-1\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)

Vậy...

\(B=\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)

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

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

\(\left(5+2\sqrt{6}\right)\left(49+20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}=\left(5+2\sqrt{6}\right)^3\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\left(\sqrt{3}+\sqrt{2}\right)^6\left(\sqrt{3}-\sqrt{2}\right)=\left(\sqrt{3}+\sqrt{2}\right)^5.1=\left(\sqrt{3}+\sqrt{2}\right)^5\)

$\left(5+2\sqrt{6}\right)\left(49+20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}

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

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

=\left(\sqrt{3}+\sqrt{2}\right)^5.1

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

\(\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}\)

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

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

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

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

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

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

\(=\left(\sqrt{3}+\sqrt{2}\right)\left(49-20\sqrt{6}\right)\)

\(\)

\(=\left(2+2\sqrt{2}\cdot\sqrt{3}+3\right)\left(25-2\cdot5\cdot\sqrt{24}+24\right)\sqrt{3-2\sqrt{3}\sqrt{2}+2}\)

\(=\left(\sqrt{2}+\sqrt{3}\right)^2\left(5-\sqrt{24}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

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

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

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

\(=9\sqrt{3}-11\sqrt{2}\)

\(pt\Leftrightarrow\left(5-2\sqrt{6}\right)^{\frac{x}{2}}+\left(5+2\sqrt{6}\right)^{\frac{x}{2}}=10\)

Thấy rằng \(5-2\sqrt{6}\) là nghịch đảo của \(5+2\sqrt{6}\), Vì vậy 

\(\left(5-2\sqrt{6}\right)^{\frac{x}{2}}\left(5+2\sqrt{6}\right)^{\frac{x}{2}}=1\)

Đặt \(\left(5-2\sqrt{6}\right)^{\frac{x}{2}}=t\) ta dc pt sau 

\(t+\frac{1}{t}=10\Rightarrow t^2-10t+1=0\Rightarrow t=5\pm2\sqrt{6}\)

Vì vậy \(t=5\pm2\sqrt{6}=\left(5-2\sqrt{6}\right)^{\pm1}=\left(5-2\sqrt{6}\right)^{\frac{x}{2}}\)

Suy ra \(\frac{x}{2}=\pm1\Rightarrow x=\pm2\) 

Bạn tham khảo 

https://hoc24.vn/cau-hoi/rut-gonfracleft52sqrt6rightleft49-20sqrt6rightsqrt5-2sqrt69sqrt3-11sqrt2.227145517764