Đáp án:
\(\begin{array}{l}
14)A = 2x - 2\\
15)A = - \dfrac{{\sqrt x - 2}}{x}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
14)A = \left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x + 1}} + \sqrt x } \right].\left[ {\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x - 1}} - \sqrt x } \right] - {x^2} + 2x - 1\\
= \left( {x - \sqrt x + 1 + \sqrt x } \right).\left( {x + \sqrt x + 1 - \sqrt x } \right) - {x^2} + 2x - 1\\
= \left( {x + 1} \right)\left( {x - 1} \right) - {x^2} + 2x - 1\\
= {x^2} - 1 - {x^2} + 2x - 1\\
= 2x - 2\\
15)A = \left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} - \dfrac{{x - 1}}{{\sqrt x - 1}}} \right]:\dfrac{{\sqrt x - x - \sqrt x }}{{ - \left( {\sqrt x - 1} \right)}}\\
= \left[ {\dfrac{{x - \sqrt x + 1 - x + 1}}{{\sqrt x - 1}}} \right].\dfrac{{ - \left( {\sqrt x - 1} \right)}}{{ - x}}\\
= \dfrac{{ - \sqrt x + 2}}{{\sqrt x - 1}}.\dfrac{{\sqrt x - 1}}{x}\\
= - \dfrac{{\sqrt x - 2}}{x}
\end{array}\)
