Đáp án:
\(\begin{array}{l}
1)H = - 2\sqrt 2 \\
2)a)x = \pm 2\sqrt 2 \\
b)x = - \dfrac{3}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)H = \sqrt {3 - 2\sqrt 3 .\sqrt 2 + 2} - \left| {\sqrt 2 + \sqrt 3 } \right|\\
= \sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} - \sqrt 3 - \sqrt 2 \\
= \sqrt 3 - \sqrt 2 - \sqrt 3 - \sqrt 2 \\
= - 2\sqrt 2 \\
2)a)29 - 9\sqrt {{x^2} + 1} = 2\\
\to 9\sqrt {{x^2} + 1} = 27\\
\to \sqrt {{x^2} + 1} = 3\\
\to {x^2} + 1 = 9\\
\to {x^2} = 8\\
\to \left| x \right| = 2\sqrt 2 \\
\to x = \pm 2\sqrt 2 \\
b)\sqrt {{{\left( {x - 2} \right)}^2}} = x + 5\\
\to \left| {x - 2} \right| = x + 5\\
\to \left[ \begin{array}{l}
x - 2 = x + 5\left( {DK:x \ge 2} \right)\\
x - 2 = - x - 5\left( {DK:x < 2} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 2 = 5\left( l \right)\\
2x = - 3
\end{array} \right.\\
\to x = - \dfrac{3}{2}
\end{array}\)
