Đáp án:
\(\begin{array}{l}
B1:\\
a)\left[ \begin{array}{l}
x = 9\\
x = 1
\end{array} \right.\\
b)x = 6\\
c)x = \dfrac{{169}}{{144}}\\
B2:\\
A = - 4\\
B = - 2022
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)\sqrt {{{\left( {x - 5} \right)}^2}} = 4\\
\to \left| {x - 5} \right| = 4\\
\to \left[ \begin{array}{l}
x - 5 = 4\\
x - 5 = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = 1
\end{array} \right.\\
b)\sqrt {4{x^2} + 4x + 1} = 3x - 5\\
\to \sqrt {{{\left( {2x + 1} \right)}^2}} = 3x - 5\\
\to \left| {2x + 1} \right| = 3x - 5\\
\to \left[ \begin{array}{l}
2x + 1 = 3x - 5\left( {DK:x \ge - \dfrac{1}{2}} \right)\\
2x + 1 = - 3x + 5\left( {DK:x < - \dfrac{1}{2}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 6\\
5x = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 6\\
x = \dfrac{4}{5}\left( l \right)
\end{array} \right.\\
\to x = 6\\
c)DK:x \ge 1\\
\sqrt {x - 1} + \sqrt {4\left( {x - 1} \right)} + 3\sqrt {9\left( {x - 1} \right)} = 5\\
\to \sqrt {x - 1} + 2\sqrt {x - 1} + 3.3.\sqrt {x - 1} = 5\\
\to 12\sqrt {x - 1} = 5\\
\to \sqrt {x - 1} = \dfrac{5}{{12}}\\
\to x - 1 = \dfrac{{25}}{{144}}\\
\to x = \dfrac{{169}}{{144}}\\
B2:\\
A = \sqrt {5 - 2.2.\sqrt 5 + 4} - \sqrt {5 + 2.2.\sqrt 5 + 4} \\
= \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 + 2} \right)}^2}} \\
= \sqrt 5 - 2 - \sqrt 5 - 2\\
= - 4\\
B = \dfrac{{\left( {\sqrt x - 2} \right)\left( {x + 2\sqrt x + 4} \right)}}{{x + 2\sqrt x + 4}} - \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x - 1}} - 2021\\
= \sqrt x - 2 - \sqrt x + 1 - 2021\\
= - 2022
\end{array}\)
