$A = \left( {\dfrac{{\sqrt x }}{4} - \dfrac{1}{{4\sqrt x }}} \right)\left( {\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}} \right)$
ĐKXĐ:$x \ne 0;x \ge 1$
a,$A = \dfrac{{x - 1}}{{4\sqrt x }}.\dfrac{{{{(\sqrt x - 1)}^2} - {{(\sqrt x + 1)}^2}}}{{x - 1}} = \dfrac{{x - 2\sqrt x + 1 - x - 2\sqrt x - 1}}{{4\sqrt x }} = \dfrac{{ - 4\sqrt x }}{{4\sqrt x }} = - 1$
b, Để $2A + \sqrt x = \dfrac{5}{4}$
=>$ - 2 + \sqrt x = \dfrac{5}{4} \Leftrightarrow \sqrt x = \dfrac{{13}}{4} \Rightarrow x = \dfrac{{169}}{{16}}$
- maikhai yêu thích