Đáp án:

$1)A = 8\sqrt 2 ;B =  - 4\sqrt {x + 1} $

$2)a) $ Không tồn tại $x$; $b)x=1$; $c)x=3$

$3)$$a)\dfrac{{\sqrt {10} }}{5};b)\sqrt a $

Giải thích các bước giải:

$\begin{array}{l}
1)\\
 + )A = \sqrt {32}  + \sqrt {50}  - 2\sqrt 8  + \sqrt {18} \\
 = \sqrt {16.2}  + \sqrt {25.2}  - 2.\sqrt {4.2}  + \sqrt {9.2} \\
 = 4\sqrt 2  + 5\sqrt 2  - 2.2.\sqrt 2  + 3\sqrt 2 \\
 = 8\sqrt 2 \\
 + )B = \sqrt {16x + 16}  - 9\sqrt {x + 1}  + \dfrac{1}{5}\sqrt {25x + 25} \left( {DK:x \ge  - 1} \right)\\
 = \sqrt {16\left( {x + 1} \right)}  - 9\sqrt {x + 1}  + \dfrac{1}{5}\sqrt {25\left( {x + 1} \right)} \\
 = 4\sqrt {x + 1}  - 9\sqrt {x + 1}  + \dfrac{1}{5}.5\sqrt {x + 1} \\
 =  - 4\sqrt {x + 1} 
\end{array}$

$2)$

a) Ta có:

$\sqrt {4x - 20}  - 3\sqrt {\dfrac{{x - 5}}{9}}  = \sqrt {1 - x} $ có nghĩa

$ \Leftrightarrow \left\{ \begin{array}{l}
4x - 20 \ge 0\\
x - 5 \ge 0\\
1 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 5\\
x \le 1
\end{array} \right. \Leftrightarrow 5 \le x \le 1\left( {mt} \right)$

b)${DK:x \ge \dfrac{1}{2}}$

Ta có:

$\begin{array}{l}
\sqrt {50x - 25}  + \sqrt {8x - 4}  - 3\sqrt x  = \sqrt {72x - 36}  - \sqrt {4x} \\
 \Leftrightarrow \sqrt {25\left( {2x - 1} \right)}  + \sqrt {4\left( {2x - 1} \right)}  - 3\sqrt x  = \sqrt {36\left( {2x - 1} \right)}  - 2\sqrt x \\
 \Leftrightarrow 5\sqrt {2x - 1}  + 2\sqrt {2x - 1}  - 3\sqrt x  = 6\sqrt {2x - 1}  - 2\sqrt x \\
 \Leftrightarrow 5\sqrt {2x - 1}  + 2\sqrt {2x - 1}  - 6\sqrt {2x - 1}  = 3\sqrt x  - 2\sqrt x \\
 \Leftrightarrow \sqrt {2x - 1}  = \sqrt x \\
 \Leftrightarrow 2x - 1 = x\\
 \Leftrightarrow x = 1\left( {tm} \right)
\end{array}$

c) ${DK:x \ge 3}$

Ta có:

$\begin{array}{l}
\sqrt {{x^2} - 9}  - \sqrt {4x - 12}  = 0\\
 \Leftrightarrow \sqrt {\left( {x - 3} \right)\left( {x + 3} \right)}  - \sqrt {4\left( {x - 3} \right)}  = 0\\
 \Leftrightarrow \sqrt {x - 3} .\sqrt {x + 3}  - 2\sqrt {x - 3}  = 0\\
 \Leftrightarrow \sqrt {x - 3} \left( {\sqrt {x + 3}  - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 3}  = 0\\
\sqrt {x + 3}  - 2 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x + 3 = 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\left( c \right)\\
x = 1\left( l \right)
\end{array} \right.\\
 \Leftrightarrow x = 3
\end{array}$

$3)$

$a)\sqrt {\dfrac{2}{5}}  = \dfrac{{\sqrt 2 }}{{\sqrt 5 }} = \dfrac{{\sqrt 2 .\sqrt 5 }}{{\sqrt 5 .\sqrt 5 }} = \dfrac{{\sqrt {10} }}{5}$

$\begin{array}{l}
b)DK:a > 0\\
a\sqrt {\dfrac{1}{a}}  = \dfrac{a}{{\sqrt a }} = \sqrt a 
\end{array}$