Đáp án:
\(\begin{array}{l}
E = \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
K = \sqrt x - 1\\
T = \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x }}\\
R = - 2\sqrt x
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
E = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) - x}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}:\dfrac{{\sqrt x \left( {\sqrt x + 1} \right) - \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{x + \sqrt x - x}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{{x + \sqrt x - \sqrt x }}\\
= \dfrac{{\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\dfrac{{\sqrt x + 1}}{x}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
K = \dfrac{{x\sqrt x - 2x + \sqrt x }}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x \left( {x - 2\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x {{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \sqrt x - 1\\
T = \dfrac{{x - 1}}{{\sqrt x }}:\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - 1}}{{\sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x - 1 + 1 - \sqrt x }}\\
= \dfrac{{x - 1}}{{\sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x - \sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x }}\\
R = \dfrac{{x - 1}}{{2\sqrt x }}.\dfrac{{\left( {x - \sqrt x } \right)\left( {\sqrt x - 1} \right) - \left( {x + \sqrt x } \right)\left( {\sqrt x + 1} \right)}}{{x - 1}}\\
= \dfrac{{x\sqrt x - 2x + \sqrt x - x\sqrt x - 2x - \sqrt x }}{{2\sqrt x }}\\
= \dfrac{{ - 4x}}{{2\sqrt x }}\\
= - 2\sqrt x
\end{array}\)
