Đáp án:
\(\begin{array}{l}
C17:\\
b)A = \dfrac{{\sqrt 5 }}{{3 + \sqrt 5 }}\\
C18:\\
a)A = \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}\\
b)M = \dfrac{{\sqrt 5 }}{{\sqrt 5 - 1}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C17:\\
b)Thay:x = 6 + 2\sqrt 5 = 5 + 2\sqrt 5 .1 + 1\\
= {\left( {\sqrt 5 + 1} \right)^2}\\
\to A = \dfrac{{\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - 1}}{{2 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }} = \dfrac{{\sqrt 5 + 1 - 1}}{{2 + \sqrt 5 + 1}}\\
= \dfrac{{\sqrt 5 }}{{3 + \sqrt 5 }}\\
C18:\\
a)A = \dfrac{{\sqrt x \left( {\sqrt x - 2} \right) + 3\left( {\sqrt x + 2} \right) + 3\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}:\dfrac{{x + 3\sqrt x - 2\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 2\sqrt x + 3\sqrt x + 6 + 3\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{x + \sqrt x }}\\
= \dfrac{{x + 4\sqrt x + 4}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{x + \sqrt x }}\\
= \dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}\\
b)Thay:x = 9 - 4\sqrt 5 \\
= 5 - 2.2.\sqrt 5 + 4\\
= {\left( {\sqrt 5 - 2} \right)^2}\\
\to M = \dfrac{{\sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} + 2}}{{\sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} + 1}} = \dfrac{{\sqrt 5 - 2 + 2}}{{\sqrt 5 - 2 + 1}}\\
= \dfrac{{\sqrt 5 }}{{\sqrt 5 - 1}}
\end{array}\)
