Đáp án:
\(\begin{array}{l}
a)\dfrac{2}{{x + \sqrt x + 1}}\\
b)Q = \dfrac{2}{{8 - \sqrt 7 }}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
Q = \dfrac{{x + 2 + \sqrt x \left( {\sqrt x - 1} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}:\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{x + 2 + x - \sqrt x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{2}{{x + \sqrt x + 1}}\\
b)Thay:x = 8 - 2\sqrt 7 = 7 - 2\sqrt 7 .1 + 1\\
= {\left( {\sqrt 7 - 1} \right)^2}\\
\to Q = \dfrac{2}{{8 - 2\sqrt 7 + \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} + 1}}\\
= \dfrac{2}{{9 - 2\sqrt 7 + \sqrt 7 - 1}}\\
= \dfrac{2}{{8 - \sqrt 7 }}
\end{array}\)
