Đáp án:
\(\begin{array}{l}
a)\dfrac{{x + 4}}{{4\sqrt x }}\\
b)x = 4\\
c)\dfrac{{6 - \sqrt 7 }}{{2\left( {\sqrt 7 - 1} \right)}}\\
d)0 < x < 4\\
e)Min = 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x > 0;x \ne 1\\
P = \dfrac{{\left( { - x + \sqrt x - 3 - \sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}:\dfrac{{x + 2\sqrt x + 1 - x + 2\sqrt x - 1 - 8\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{ - x - 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{ - 4\sqrt x }}\\
= \dfrac{{x + 4}}{{4\sqrt x }}\\
b)P = 1\\
\to \dfrac{{x + 4}}{{4\sqrt x }} = 1\\
\to x + 4 = 4\sqrt x \\
\to x - 4\sqrt x + 4 = 0\\
\to {\left( {\sqrt x - 2} \right)^2} = 0\\
\to \sqrt x - 2 = 0\\
\to x = 4\\
c)Thay:x = 8 - 2\sqrt 7 \\
= 7 - 2\sqrt 7 .1 + 1\\
= {\left( {\sqrt 7 - 1} \right)^2}\\
\to P = \dfrac{{8 - 2\sqrt 7 + 4}}{{4\sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} }}\\
= \dfrac{{12 - 2\sqrt 7 }}{{4\left( {\sqrt 7 - 1} \right)}} = \dfrac{{6 - \sqrt 7 }}{{2\left( {\sqrt 7 - 1} \right)}}\\
d)P > \dfrac{{\sqrt x + 2}}{4}\\
\to \dfrac{{x + 4}}{{4\sqrt x }} > \dfrac{{\sqrt x + 2}}{4}\\
\to \dfrac{{x + 4 - x - 2\sqrt x }}{{4\sqrt x }} > 0\\
\to 4 - 2\sqrt x > 0\left( {do:4\sqrt x > 0\forall x > 0} \right)\\
\to 2 > \sqrt x \\
\to 0 < x < 4\\
e)P = \dfrac{{x + 4}}{{4\sqrt x }} = \dfrac{{\sqrt x }}{4} + \dfrac{1}{{\sqrt x }}\\
Do:x > 0\\
BDT:Co - si:\dfrac{{\sqrt x }}{4} + \dfrac{1}{{\sqrt x }} \ge 2\sqrt {\dfrac{{\sqrt x }}{4}.\dfrac{1}{{\sqrt x }}} = 1\\
\to Min = 1\\
\Leftrightarrow \dfrac{{\sqrt x }}{4} = \dfrac{1}{{\sqrt x }}\\
\to x = 4
\end{array}\)
