Đáp án:
d) x=-4
e) x=9
Giải thích các bước giải:
\(\begin{array}{l}
d)DK:3 \ge x\\
\sqrt {3 - x} - \sqrt {{x^2} - 9} = 0\\
\to \sqrt {3 - x} = \sqrt {{x^2} - 9} \\
\to 3 - x = {x^2} - 9\\
\to {x^2} + x - 12 = 0\\
\to \left( {x + 4} \right)\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 4\\
x = 3\left( l \right)
\end{array} \right.\\
e)DK:x \ge 0\\
\dfrac{{x - \sqrt x }}{{\sqrt x + 1}} = \dfrac{3}{2}\\
\to 2x - 2\sqrt x = 3\sqrt x + 3\\
\to 2x - 5\sqrt x - 3 = 0\\
\to \left( {\sqrt x - 3} \right)\left( {2\sqrt x + 1} \right) = 0\\
\to \sqrt x - 3 = 0\left( {do:2\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to x = 9
\end{array}\)
