Đáp án:
\(\begin{array}{l}
a)x = 1\\
b)x = 1\\
c)x = \dfrac{1}{4}\\
d)0 \le x < 1\\
e)0 \le x < 9
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\sqrt x + 1 = 2\sqrt x \\
\to \sqrt x = 1\\
\to x = 1\\
b)\dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}} = \dfrac{3}{2}\\
\to 4\sqrt x + 2 = 3\sqrt x + 3\\
\to \sqrt x = 1\\
\to x = 1\\
c)\dfrac{{\sqrt x - 2}}{{\sqrt x - 3}} = \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
\to \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right) = \left( {\sqrt x - 3} \right)\left( {\sqrt x + 1} \right)\\
\to x - 4 = x - 2\sqrt x - 3\\
\to 2\sqrt x = 1\\
\to \sqrt x = \dfrac{1}{2}\\
\to x = \dfrac{1}{4}\\
d)\dfrac{1}{{\sqrt x - 1}} < 0\\
\to \sqrt x - 1 < 0\\
\to 0 \le x < 1\\
e)\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} < \dfrac{1}{2}\\
\to \dfrac{{2\sqrt x - 2 - \sqrt x - 1}}{{2\left( {\sqrt x + 1} \right)}} < 0\\
\to \sqrt x - 3 < 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0\forall x \ge 0} \right)\\
\to 0 \le x < 9
\end{array}\)
