Giúp mình với cảm ơn trước ạ

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
\sqrt {81}  - \sqrt {80} .\sqrt {0,2}  = \sqrt {81}  - \sqrt {80.0,2}  = \sqrt {81}  - \sqrt {16}  = \sqrt {{9^2}}  - \sqrt {{4^2}}  = 9 - 4 = 5\\
b,\\
\sqrt {{{\left( {2 - \sqrt 5 } \right)}^2}}  - \dfrac{1}{2}\sqrt {20}  = \left| {2 - \sqrt 5 } \right| - \dfrac{1}{2}\sqrt {{2^2}.5}  = \left( {\sqrt 5  - 2} \right) - \dfrac{1}{2}.2.\sqrt 5  = \sqrt 5  - 2 - \sqrt 5  =  - 2\\
2,\\
a,\,\,\,\,\,\,\,\,\,\,\sqrt { - x + 1} \\
DKXD:\,\,\,\, - x + 1 \ge 0 \Leftrightarrow  - \left( {x - 1} \right) \ge 0 \Leftrightarrow x - 1 \le 0 \Leftrightarrow x \le 1\\
b,\,\,\,\,\,\,\,\sqrt {\dfrac{1}{{{x^2} - 2x + 1}}} \\
DKXD:\,\,\,\,\left\{ \begin{array}{l}
\dfrac{1}{{{x^2} - 2x + 1}} \ge 0\\
{x^2} - 2x + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{{{{\left( {x - 1} \right)}^2}}} \ge 0\\
{\left( {x - 1} \right)^2} \ne 0
\end{array} \right. \Leftrightarrow x \ne 1\\
2,\\
1,\\
a,\\
ab + b\sqrt a  + \sqrt a  + 1\\
 = \left( {ab + b\sqrt a } \right) + \left( {\sqrt a  + 1} \right)\\
 = b\sqrt a .\left( {\sqrt a  + 1} \right) + \left( {\sqrt a  + 1} \right)\\
 = \left( {\sqrt a  + 1} \right)\left( {b\sqrt a  + 1} \right)\\
2,\\
\sqrt {9x + 9}  + \sqrt {x + 1}  = 20\,\,\,\,\,\,\,\,\left( {x \ge  - 1} \right)\\
 \Leftrightarrow \sqrt {9\left( {x + 1} \right)}  + \sqrt {x + 1}  = 20\\
 \Leftrightarrow 3.\sqrt {x + 1}  + \sqrt {x + 1}  = 20\\
 \Leftrightarrow 4\sqrt {x + 1}  = 20\\
 \Leftrightarrow \sqrt {x + 1}  = 5\\
 \Leftrightarrow x + 1 = 25\\
 \Leftrightarrow x = 24
\end{array}\)