Đáp án:
a. \(\dfrac{{ - \sqrt x - 2}}{{x + \sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 1\\
P = \dfrac{{x + 2 - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{x + 2 - x + 1 - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{ - x - \sqrt x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\left( {1 - \sqrt x } \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{ - \sqrt x - 2}}{{x + \sqrt x + 1}}\\
b.Thay:x = 33 - 8\sqrt 2 \\
\to P = \dfrac{{ - \sqrt {33 - 8\sqrt 2 } - 2}}{{33 - 8\sqrt 2 + \sqrt {33 - 8\sqrt 2 } + 1}}\\
= \dfrac{{ - \sqrt {33 - 8\sqrt 2 } - 2}}{{\sqrt {33 - 8\sqrt 2 } + 34 - 8\sqrt 2 }}
\end{array}\)
