Đáp án:
\(\begin{array}{l}
a)\dfrac{{9 - \sqrt 3 }}{{13}}\\
b)\dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
c)0 \le x < \dfrac{1}{{16}}\\
d)Min = - \dfrac{1}{2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 7 - 4\sqrt 3 \\
= 4 - 2.2.\sqrt 3 + 3\\
= {\left( {2 - \sqrt 3 } \right)^2}\\
\to Q = \dfrac{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + 1}}{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} + 2}}\\
= \dfrac{{2 - \sqrt 3 + 1}}{{2 - \sqrt 3 + 2}} = \dfrac{{3 - \sqrt 3 }}{{4 - \sqrt 3 }} = \dfrac{{9 - \sqrt 3 }}{{13}}\\
b)P = \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + \sqrt x + 1 - 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - \sqrt x - \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
M = P.Q = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}.\dfrac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}}\\
c)M < - \dfrac{1}{3}\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} < - \dfrac{1}{3}\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} + \dfrac{1}{3} < 0\\
\to \dfrac{{3\sqrt x - 3 + \sqrt x + 2}}{{3\left( {\sqrt x + 2} \right)}} < 0\\
\to 4\sqrt x - 1 < 0\left( {do:\sqrt x + 2 > 0\forall x \ge 0} \right)\\
\to 0 \le x < \dfrac{1}{{16}}\\
d)M = \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 - 3}}{{\sqrt x + 2}}\\
= 1 - \dfrac{3}{{\sqrt x + 2}}\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 2 \ge 2\\
\to \dfrac{3}{{\sqrt x + 2}} \le \dfrac{3}{2}\\
\to - \dfrac{3}{{\sqrt x + 2}} \ge - \dfrac{3}{2}\\
\to 1 - \dfrac{3}{{\sqrt x + 2}} \ge - \dfrac{1}{2}\\
\to Min = - \dfrac{1}{2}\\
\Leftrightarrow x = 0
\end{array}\)
