Đáp án:
c) x>4
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x > 0;x \ne 4\\
b)P = \left[ {\dfrac{{4\sqrt x \left( {\sqrt x - 2} \right) - 8x}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right]:\left[ {\dfrac{{\sqrt x - 2 - 2\left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right]\\
= \dfrac{{4x - 8\sqrt x - 8x}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x - 2 - 2\sqrt x + 4}}\\
= \dfrac{{ - 4x - 8\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x - 2 - 2\sqrt x + 4}}\\
= \dfrac{{ - 4\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{ - \sqrt x + 2}}\\
= \dfrac{{4\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x - 2}}\\
= \dfrac{{4\sqrt x }}{{\sqrt x - 2}}\\
c)P > 1\\
\to \dfrac{{4\sqrt x }}{{\sqrt x - 2}} > 1\\
\to \dfrac{{4\sqrt x - \sqrt x + 2}}{{\sqrt x - 2}} > 0\\
\to \dfrac{{3\sqrt x + 2}}{{\sqrt x - 2}} > 0\\
\to \sqrt x - 2 > 0\left( {do:3\sqrt x + 2 > 0\forall x} \right)\\
\to x > 4
\end{array}\)
