Đáp án:
b. \(\dfrac{1}{{\sqrt x }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 9\\
\dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right) - 3x + 1 - \sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x - 6\sqrt x + x + 4\sqrt x + 3 - 3x + 1 - \sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{ - 3\sqrt x + 4}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
b.DK:x > 0;x \ne 1\\
\left[ {\dfrac{{x - \sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}} \right]:\left[ {\dfrac{{\sqrt x - 1 + 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right]\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{1}{{\sqrt x }}\\
c.DK:a > 0;a \ne 1\\
\left[ {\dfrac{{\sqrt a - \sqrt a + 1}}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right]:\left( {\dfrac{{\sqrt a + 1 - \sqrt a - 2}}{{\sqrt a - 1}}} \right)\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\sqrt a - 1}}{{ - 1}}\\
= - \dfrac{1}{{\sqrt a }}
\end{array}\)
