Đáp án:
\(\begin{array}{l}
a)\left[ \begin{array}{l}
A = - 6\\
A = - 2x\\
A = 6
\end{array} \right.\\
b)x = - \dfrac{1}{2}\\
c)A = - 2\sqrt 3 + 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \sqrt {{{\left( {x - 3} \right)}^2}} - \sqrt {{{\left( {x + 3} \right)}^2}} \\
= \left| {x - 3} \right| - \left| {x + 3} \right|\\
\to \left[ \begin{array}{l}
A = x - 3 - \left( {x + 3} \right)\left( {DK:x \ge 3} \right)\\
A = - x + 3 - \left( {x + 3} \right)\left( {DK:3 > x > - 3} \right)\\
A = - x + 3 + \left( {x + 3} \right)\left( {DK: - 3 > x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = - 6\\
A = - 2x\\
A = 6
\end{array} \right.\\
b)A = 1\\
\Leftrightarrow - 2x = 1\\
\to x = - \dfrac{1}{2}\\
c)Thay:x = \sqrt {4 - 2\sqrt 3 } \\
= \sqrt {3 - 2\sqrt 3 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
= \sqrt 3 - 1\\
\to A = - 2\left( {\sqrt 3 - 1} \right) = - 2\sqrt 3 + 2
\end{array}\)
