Đáp án:
B61:
c. x>4
Giải thích các bước giải:
\(\begin{array}{l}
B60)\\
M = \left[ {\dfrac{{x - 1 - 4\sqrt x + \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right].\dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 3\sqrt x }}{{x - 1}}.\dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}} = \dfrac{{\sqrt x - 2 - 1}}{{\sqrt x - 2}} = 1 - \dfrac{1}{{\sqrt x - 2}}\\
c.M \in Z\\
\Leftrightarrow \dfrac{1}{{\sqrt x - 2}} \in Z\\
\Leftrightarrow \sqrt x - 2 \in U\left( 1 \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x - 2 = 1\\
\sqrt x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 9\\
x = 1\left( l \right)
\end{array} \right.\\
d.Thay:x = 7 - 2\sqrt 6 = 6 - 2\sqrt 6 .1 + 1\\
= {\left( {\sqrt 6 - 1} \right)^2}\\
\to M = \dfrac{{\sqrt {{{\left( {\sqrt 6 - 1} \right)}^2}} - 3}}{{\sqrt {{{\left( {\sqrt 6 - 1} \right)}^2}} - 2}} = \dfrac{{\sqrt 6 - 1 - 3}}{{\sqrt 6 - 1 - 2}}\\
= \dfrac{{\sqrt 6 - 4}}{{\sqrt 6 - 3}}\\
B61:\\
a.R = \left[ {\dfrac{{x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right].\left[ {\dfrac{{\sqrt x - 2 + 4}}{{x - 4}}} \right]\\
= \dfrac{{x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x + 2}}{{x - 4}}\\
= \dfrac{{\sqrt x + 2}}{{x - 2\sqrt x }}\\
b.Thay:x = 4 + 2\sqrt 3 = {\left( {\sqrt 3 + 1} \right)^2}\\
\to R = \dfrac{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + 2}}{{4 + 2\sqrt 3 - 2\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}\\
= \dfrac{{\sqrt 3 + 1 + 2}}{{4 + 2\sqrt 3 - 2\sqrt 3 - 2}} = \dfrac{{\sqrt 3 + 3}}{2}\\
c.R > 0\\
\to \dfrac{{\sqrt x + 2}}{{x - 2\sqrt x }} > 0\\
\to x - 2\sqrt x > 0\left( {do:\sqrt x + 2 > 0\forall x > 0} \right)\\
\to \sqrt x - 2 > 0\left( {do:\sqrt x > 0\forall x > 0} \right)\\
\to x > 4
\end{array}\)
