Đáp án:
16) \(\dfrac{3}{{\sqrt x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
11)DK:x \ge 0;x \ne 1\\
A = \left[ {\dfrac{{2\sqrt x + x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}} \right].\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{x + \sqrt x + 1}}{{\sqrt x + 2}}\\
= \dfrac{1}{{\sqrt x + 2}}\\
13)DK:x \ge 0;x \ne 1\\
A = \dfrac{{15\sqrt x - 11 - \left( {3\sqrt x - 2} \right)\left( {\sqrt x + 3} \right) - \left( {2\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{15\sqrt x - 11 - 3x - 7\sqrt x + 6 - 2x - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{ - 5x + 7\sqrt x - 2}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {2 - 5\sqrt x } \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2 - 5\sqrt x }}{{\sqrt x + 3}}\\
14)DK:x \ge 0;x \ne 1\\
A = \dfrac{{x\sqrt x + 1 - \left( {\sqrt x - 1} \right)\left( {x - 1} \right)}}{{x - 1}}\\
= \dfrac{{x\sqrt x + 1 - x\sqrt x + \sqrt x + x - 1}}{{x - 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x - 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 1}}\\
15)DK:x > 0;x \ne 1\\
A = \left[ {\dfrac{{x - 1 - 4\sqrt x + 4 + \sqrt x - 1}}{{x - 1}}} \right].\dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2}}{{x - 1}}.\dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x }}\\
16)DK:x \ge 0;x \ne 9\\
A = \dfrac{{x - 3\sqrt x + 2x + 6\sqrt x - 3x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{3\sqrt x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{3}{{\sqrt x + 3}}
\end{array}\)
