Đáp án:
\(\begin{array}{l}
a)\dfrac{{2\sqrt x + 2x + 2}}{{\sqrt x }}\\
b)x \in \emptyset
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}} + \dfrac{{x - 1}}{{\sqrt x }}.\dfrac{{{{\left( {\sqrt x + 1} \right)}^2} + {{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} - \dfrac{{x - \sqrt x + 1}}{{\sqrt x }} + \dfrac{{x + 2\sqrt x + 1 + x - 2\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1 - x + \sqrt x - 1}}{{\sqrt x }} + \dfrac{{2x + 2}}{{\sqrt x }}\\
= \dfrac{{2\sqrt x + 2x + 2}}{{\sqrt x }}\\
b)A = 6\\
\to \dfrac{{2\sqrt x + 2x + 2}}{{\sqrt x }} = 6\\
\to 2\sqrt x + 2x + 2 = 6\sqrt x \\
\to 2x - 4\sqrt x + 2 = 0\\
\to 2{\left( {\sqrt x - 1} \right)^2} = 0\\
\to \sqrt x - 1 = 0\\
\to x = 1\left( {KTM} \right)\\
\to x \in \emptyset
\end{array}\)
