Đáp án:
\(\begin{array}{l}
a)\dfrac{{x + \sqrt x }}{{3\sqrt x - 1}}\\
b)\dfrac{{7 + 3\sqrt 5 }}{{3\sqrt 5 + 2}}\\
c)x \in \emptyset
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne \dfrac{1}{9}\\
Q = \dfrac{{\left( {\sqrt x - 1} \right)\left( {3\sqrt x + 1} \right) - 3\sqrt x + 1 + 8\sqrt x }}{{\left( {3\sqrt x + 1} \right)\left( {3\sqrt x - 1} \right)}}:\dfrac{{3\sqrt x + 1 - 3\sqrt x + 2}}{{3\sqrt x + 1}}\\
= \dfrac{{3x - 2\sqrt x - 1 + 5\sqrt x + 1}}{{\left( {3\sqrt x + 1} \right)\left( {3\sqrt x - 1} \right)}}.\dfrac{{3\sqrt x + 1}}{3}\\
= \dfrac{{3x + 3\sqrt x }}{{\left( {3\sqrt x + 1} \right)\left( {3\sqrt x - 1} \right)}}.\dfrac{{3\sqrt x + 1}}{3}\\
= \dfrac{{x + \sqrt x }}{{3\sqrt x - 1}}\\
b)Thay:x = 6 + 2\sqrt 5 = 5 + 2\sqrt 5 .1 + 1\\
= {\left( {\sqrt 5 + 1} \right)^2}\\
\to Q = \dfrac{{6 + 2\sqrt 5 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }}{{3\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - 1}} = \dfrac{{6 + 2\sqrt 5 + \sqrt 5 + 1}}{{3\left( {\sqrt 5 + 1} \right) - 1}}\\
= \dfrac{{7 + 3\sqrt 5 }}{{3\sqrt 5 + 2}}\\
c)Q = \dfrac{5}{6}\\
\to \dfrac{{x + \sqrt x }}{{3\sqrt x - 1}} = \dfrac{5}{6}\\
\to 6x + 6\sqrt x = 15\sqrt x - 5\\
\to 6x - 9\sqrt x + 5 = 0\left( {vô nghiệm} \right)\\
\to x \in \emptyset
\end{array}\)
