Đáp án:
B2:
b) x=0
Giải thích các bước giải:
\(\begin{array}{l}
1c)\sqrt {9 - 2.3.\sqrt 5 + 5} + 7\\
= \sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} + 7\\
= 3 - \sqrt 5 + 7 = 10 - \sqrt 5 \\
B2:\\
a)DK:x - 6 \ge 0 \to x \ge 6\\
b)P = \dfrac{2}{{\sqrt x - 2}}:\left[ {\dfrac{{\sqrt x + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right]\\
= \dfrac{2}{{\sqrt x - 2}}:\dfrac{{2\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{2}{{\sqrt x - 2}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{2\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}\\
P = \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 + 1}}{{\sqrt x + 1}} = 1 + \dfrac{1}{{\sqrt x + 1}}\\
P \in Z \Leftrightarrow \dfrac{1}{{\sqrt x + 1}} \in Z\\
\Leftrightarrow \sqrt x + 1 \in U\left( 1 \right)\\
\to \sqrt x + 1 = 1\\
\to x = 0
\end{array}\)
