Đáp án:
\(\begin{array}{l}
a,\\
Q = \dfrac{{x + 2\sqrt x + 1}}{{\sqrt x }}\\
b,\\
Q > 4,\,\,\,\forall x > 0,x \ne 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
DKXD:\,\,\,\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.\\
a,\\
Q = \dfrac{{x\sqrt x - 1}}{{x - \sqrt x }} - \dfrac{{x\sqrt x + 1}}{{x + \sqrt x }} + \dfrac{{x + 1}}{{\sqrt x }}\\
= \dfrac{{{{\sqrt x }^3} - {1^3}}}{{{{\sqrt x }^2} - \sqrt x }} - \dfrac{{{{\sqrt x }^3} + {1^3}}}{{{{\sqrt x }^2} + \sqrt x }} + \dfrac{{x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x - 1} \right).\left( {{{\sqrt x }^2} + \sqrt x .1 + {1^2}} \right)}}{{\sqrt x .\left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x + 1} \right).\left( {{{\sqrt x }^2} - \sqrt x .1 + {1^2}} \right)}}{{\sqrt x .\left( {\sqrt x + 1} \right)}} + \dfrac{{x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x .\left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x .\left( {\sqrt x + 1} \right)}} + \dfrac{{x + 1}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} - \dfrac{{x - \sqrt x + 1}}{{\sqrt x }} + \dfrac{{x + 1}}{{\sqrt x }}\\
= \dfrac{{\left( {x + \sqrt x + 1} \right) - \left( {x - \sqrt x + 1} \right) + \left( {x + 1} \right)}}{{\sqrt x }}\\
= \dfrac{{x + \sqrt x + 1 - x + \sqrt x - 1 + x + 1}}{{\sqrt x }}\\
= \dfrac{{x + 2\sqrt x + 1}}{{\sqrt x }}\\
b,\\
Q - 4 = \dfrac{{x + 2\sqrt x + 1}}{{\sqrt x }} - 4\\
= \dfrac{{x + 2\sqrt x + 1 - 4\sqrt x }}{{\sqrt x }}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\sqrt x }}\\
= \dfrac{{{{\sqrt x }^2} - 2.\sqrt x .1 + {1^2}}}{{\sqrt x }}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} > 0,\,\,\,\forall x > 0,x \ne 1\\
\Rightarrow Q - 4 > 0,\,\,\,\,\forall x > 0,x \ne 1\\
\Rightarrow Q > 4,\,\,\,\forall x > 0,x \ne 1
\end{array}\)
