Đáp án:
c) x=1
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 4\\
Q = \dfrac{{2\left( {\sqrt x - 2} \right) - \sqrt x - 2 + 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2\sqrt x - 4 + \sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3\sqrt x - 6}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{3}{{\sqrt x + 2}}\\
b)Q = \dfrac{6}{5}\\
\to \dfrac{3}{{\sqrt x + 2}} = \dfrac{6}{5}\\
\to 6\sqrt x + 12 = 15\\
\to 6\sqrt x = 3\\
\to \sqrt x = \dfrac{1}{2}\\
\to x = \dfrac{1}{4}\\
c)Q = \dfrac{3}{{\sqrt x + 2}}\\
Q \in Z\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 2}} \in Z\\
\to \sqrt x + 2 \in U\left( 3 \right)\\
Mà:\sqrt x + 2 \ge 2\forall x \ge 0\\
\to \sqrt x + 2 = 3\\
\to \sqrt x = 1\\
\to x = 1
\end{array}\)
