Đáp án:
c) x=0
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = \dfrac{{81}}{{100}}\\
\to B = \dfrac{{\sqrt {\dfrac{{81}}{{100}}} - 2}}{{\sqrt {\dfrac{{81}}{{100}}} + 1}} = \dfrac{{\dfrac{9}{{10}} - 2}}{{\dfrac{9}{{10}} + 1}} = - \dfrac{{11}}{{19}}\\
b)A = \dfrac{{\sqrt x - 2 + x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
P = A.B = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 2}}\\
c)P = \dfrac{{\sqrt x }}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 - 2}}{{\sqrt x + 2}} = 1 - \dfrac{2}{{\sqrt x + 2}}\\
P \in Z\\
\Leftrightarrow \dfrac{2}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \sqrt x + 2 \in U\left( 2 \right)\\
Mà:\sqrt x + 2 \ge 2\\
\to \sqrt x + 2 = 2\\
\to x = 0
\end{array}\)
