Đáp án:

 b) $S=∅$

 d) $S=\left\{\dfrac{17}{9}\right\}$

Giải thích các bước giải:

b) ĐKXĐ: $x\ge 3$

$\dfrac{1}{3}\sqrt{9x-27}+\sqrt{\dfrac{x-3}{4}}+2=0$

Ta có: $\begin{cases}\dfrac{1}{3}\sqrt{9x-27}\ge 0\\\sqrt{\dfrac{x-3}{4}}\ge 0\end{cases}$

$⇒\dfrac{1}{3}\sqrt{9x-27}+\sqrt{\dfrac{x-3}{4}}\ge 0$

$⇒\dfrac{1}{3}\sqrt{9x-27}+\sqrt{\dfrac{x-3}{4}}+2>0$

$⇒VT>VP$

Vậy $S=∅$.

d) ĐKXĐ: $x\le 2$

$2\sqrt{\dfrac{2-x}{4}}+\sqrt{8-4x}=1$

$⇔\sqrt{2-x}+2\sqrt{2-x}=1$

$⇔3\sqrt{2-x}=1$

$⇔9(2-x)=1$

$⇔2-x=\dfrac{1}{9}$

$⇔x=\dfrac{17}{9}\,(TM)$

Vậy $S=\left\{\dfrac{17}{9}\right\}$.

a) Điều kiện xác định $x\ge 3$

$\begin{array}{l}
b)\dfrac{1}{3}\sqrt {9x - 27}  + \sqrt {\dfrac{{x - 3}}{4}}  + 2 = 0\\
 \Leftrightarrow \dfrac{1}{3}\sqrt {9\left( {x - 3} \right)}  + \dfrac{{\sqrt {x - 3} }}{2} + 2 = 0\\
 \Leftrightarrow \dfrac{1}{3}.3\sqrt {x - 3}  + \dfrac{{\sqrt {x - 3} }}{2} + 2 = 0\\
 \Leftrightarrow \sqrt {x - 3}  + \dfrac{{\sqrt {x - 3} }}{2} + 2 = 0 \Leftrightarrow \dfrac{{3\sqrt {x - 3} }}{2} + 2 = 0\\
\dfrac{{3\sqrt {x - 3} }}{2} \ge 0 \Rightarrow \dfrac{{3\sqrt {x - 3} }}{2} + 2 \ge 2 > 0\\
 \Rightarrow PTVN\\
 \Rightarrow S = \emptyset 
\end{array}$

d) Điều kiện xác định $x\le 2$

$\begin{array}{l} 2\sqrt {\dfrac{{2 - x}}{4}}  + \sqrt {8 - 4x}  = 1\\  \Leftrightarrow 2.\dfrac{{\sqrt {2 - x} }}{2}.\sqrt {4\left( {2 - x} \right)}  = 1\\  \Leftrightarrow \sqrt {2 - x}  + 2\sqrt {2 - x}  = 1\\  \Leftrightarrow 3\sqrt {2 - x}  = 1\\  \Leftrightarrow \sqrt {2 - x}  = \dfrac{1}{3} \Leftrightarrow 2 - x = \dfrac{1}{9} \Leftrightarrow x = \dfrac{{17}}{9}\\  \Rightarrow S = \left\{ {\dfrac{{17}}{9}} \right\} \end{array}$