Đáp án:
\(\begin{array}{l}
a)1\\
b)\dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}}\\
c)x \in \emptyset \\
d)x = \dfrac{{16}}{{25}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Q = \dfrac{{\sqrt x - \sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{1}{{\sqrt x - 1}}\\
Thay:x = 4\\
\to Q = \dfrac{1}{{\sqrt 4 - 1}} = \dfrac{1}{{2 - 1}} = 1\\
b)P = \dfrac{{\sqrt x + 1 + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
M = P:Q = \dfrac{{2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}:\dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}}\\
c)M < \dfrac{3}{2}\\
\to \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}} < \dfrac{3}{2}\\
\to \dfrac{{4\sqrt x + 2 - 3\sqrt x + 3}}{{2\left( {\sqrt x + 1} \right)}} < 0\\
\to \dfrac{{\sqrt x + 5}}{{2\left( {\sqrt x + 1} \right)}} < 0\left( {KTM} \right)\\
Do:x \ge 0 \to \left\{ \begin{array}{l}
\sqrt x + 5 > 0\\
\sqrt x + 1 > 0
\end{array} \right.\\
KL:x \in \emptyset \\
d)M = \dfrac{1}{3}\\
\to \dfrac{{2\sqrt x + 1}}{{\sqrt x + 1}} = \dfrac{1}{3}\\
\to \dfrac{{6\sqrt x - 3 - \sqrt x - 1}}{{3\left( {\sqrt x + 1} \right)}} = 0\\
\to 5\sqrt x - 4 = 0\\
\to \sqrt x = \dfrac{4}{5}\\
\to x = \dfrac{{16}}{{25}}
\end{array}\)
