Đáp án:
\(\begin{array}{l}
2)x = - \dfrac{1}{2}\\
3)a)x \ge 0;x \ne 4\\
b)A = \dfrac{{\sqrt x + 1}}{{x - 4}}\\
c)x = \dfrac{{43 + 5\sqrt {61} }}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2)a)\sqrt {{{\left( {x + 3} \right)}^2}} = x - 2\\
\to \left| {x + 3} \right| = x - 2\\
\to \left[ \begin{array}{l}
x + 3 = x - 2\\
x + 3 = - x + 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
3 = - 2\left( l \right)\\
2x = - 1
\end{array} \right.\\
\to x = - \dfrac{1}{2}\\
3)a)DK:x \ge 0;x \ne 4\\
b)A = \dfrac{{\sqrt x + 2 - 1}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}} = \dfrac{{\sqrt x + 1}}{{x - 4}}\\
c)A = \dfrac{1}{5}\\
\to \dfrac{{\sqrt x + 1}}{{x - 4}} = \dfrac{1}{5}\\
\to 5\sqrt x + 5 = x - 4\\
\to x - 5\sqrt x - 9 = 0\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{{5 + \sqrt {61} }}{2}\\
\sqrt x = \dfrac{{5 - \sqrt {61} }}{2}\left( l \right)
\end{array} \right.\\
\to x = \dfrac{{43 + 5\sqrt {61} }}{2}
\end{array}\)
