Bài 1/
a/ ĐKXĐ: $x≥0$
$2\sqrt x=6\\↔\sqrt x=3\\↔x=9(TM)$
Vậy $S=\{9\}$
b/ ĐKXĐ: $\begin{cases}x^2-4\ge 0\\x-2\ge 0\end{cases}\\↔\begin{cases}\left[\begin{array}{1}x≥2\\x≤-2\end{array}\right.\\x\ge 2\end{cases}\\↔x\\ge 2$
$\sqrt{x^2-4}-\sqrt{x-2}=0\\↔\sqrt{(x-2)(x+2)}-\sqrt{x-2}=0\\↔\sqrt{x-2}(\sqrt{x+2}-1)=0\\↔\left[\begin{array}{1}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{array}\right.\\↔\left[ \begin{array}{l}x-2=0\\\sqrt{x+1}=1\end{array} \right.\\↔\left[\begin{array}{1}x=2\\x+1=1\end{array}\right.\\↔\left[\begin{array}{1}x=2(TM)\\x=0(KTM)\end{array}\right.$
Vậy $S=\{2\}$
Bài 2/
a/ ĐKXĐ: $4x^2+4x+1\ge 0\\↔(2x+1)^2\ge 0\\↔x∈\Bbb R$
$\sqrt{4x^2+4x+1}=6\\↔\sqrt{(2x+1)^2}=6\\↔|2x+1|=6\\↔\left[\begin{array}{1}2x+1=6\\2x+1=-6\end{array}\right.\\↔\left[\begin{array}{1}2x=5\\2x=-7\end{array}\right.\\↔\left[\begin{array}{1}x=\dfrac{5}{2}(TM)\\x=-\dfrac{7}{2}(TM)\end{array}\right.$
Vậy $S=\left\{\dfrac{5}{2};-\dfrac{7}{2}\right\}$
b/ ĐKXĐ: $x+5\ge 0↔x\ge -5$
$\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\\↔\sqrt{4(x+5)}+\sqrt{x+5}-\dfrac{1}{3}.\sqrt{9(x+5)}=4\\↔2\sqrt{x+5}+\sqrt{x+5}-\sqrt{x+5}=4\\↔2\sqrt{x+5}=4\\↔\sqrt{x+5}=2\\↔x+5=4\\↔x=-1(TM)$
Vậy $S=\{-1\}$
