Đáp án:
\(\begin{array}{l}
a)\dfrac{{x + 1 + \sqrt x }}{{\sqrt x - 1}}\\
b) - 5 - 2\sqrt 3 \\
c)S = - \dfrac{{8 + 5\sqrt 2 }}{2}\\
d)\left[ \begin{array}{l}
x = 16\\
x = 4\\
x = 0
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
S = \dfrac{{x + 1 + \sqrt x }}{{x + 1}}:\left[ {\dfrac{1}{{\sqrt x - 1}} - \dfrac{{2\sqrt x }}{{x\left( {\sqrt x - 1} \right) + \left( {\sqrt x - 1} \right)}}} \right]\\
= \dfrac{{x + 1 + \sqrt x }}{{x + 1}}:\dfrac{{x + 1 - 2\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{x + 1 + \sqrt x }}{{x + 1}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \dfrac{{x + 1 + \sqrt x }}{{\sqrt x - 1}}\\
b)Thay:x = 4 - 2\sqrt 3 \\
= 3 - 2\sqrt 3 .1 + 1\\
= {\left( {\sqrt 3 - 1} \right)^2}\\
\to S = \dfrac{{4 - 2\sqrt 3 + 1 + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} - 1}}\\
= \dfrac{{4 - 2\sqrt 3 + 1 + \sqrt 3 - 1}}{{\sqrt 3 - 1 - 1}}\\
= - 5 - 2\sqrt 3 \\
c)Thay:x = \dfrac{1}{2}\\
\to S = \dfrac{{\dfrac{1}{2} + 1 + \sqrt {\dfrac{1}{2}} }}{{\sqrt {\dfrac{1}{2}} - 1}} = - \dfrac{{8 + 5\sqrt 2 }}{2}\\
d)S = \dfrac{{x + 1 + \sqrt x }}{{\sqrt x - 1}}\\
= \dfrac{{x - \sqrt x + 2\sqrt x - 2 + 3}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right) + 2\left( {\sqrt x - 1} \right) + 3}}{{\sqrt x - 1}}\\
= \sqrt x + 2 + \dfrac{3}{{\sqrt x - 1}}\\
S \in Z \to \dfrac{3}{{\sqrt x - 1}} \in Z\\
\to \sqrt x - 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 3\\
\sqrt x - 1 = - 3\left( l \right)\\
\sqrt x - 1 = 1\\
\sqrt x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 16\\
x = 4\\
x = 0
\end{array} \right.
\end{array}\)
