Cho P = (1 - căn bậc hai x / (căn bậc hai x + 1)) : ( (căn bậc hai x + 3) / (căn bậc hai x

a. \(P = \left( {1 - \frac{{\sqrt x }}{{\sqrt x + 1}}} \right):\left( {\frac{{\sqrt x + 3}}{{\sqrt x - 2}} + \frac{{\sqrt x + 2}}{{3 - \sqrt x }} + \frac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}}} \right)\)

\(P = \frac{{\sqrt x + 1 - \sqrt x }}{{\sqrt x + 1}}:\left( {\frac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} - \frac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 3} \right)\left( { - 2} \right)}} + \frac{{\sqrt x + 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}} \right)\)

\(P = \frac{1}{{\sqrt x + 1}}:\frac{{x - 9 + 4 - x + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} = \frac{1}{{\sqrt x + 1}}:\frac{{\sqrt x - 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\)

\(P = \frac{1}{{\sqrt x + 1}}.\frac{{\sqrt x - 2}}{1} = \frac{{\sqrt x - 2}}{{\sqrt x + 1}}\).

b. Có: \(P = \frac{{\sqrt x - 2}}{{\sqrt x + 1}} = \frac{{\sqrt x + 1 - 3}}{{\sqrt x + 1}} = \frac{{\sqrt x + 1}}{{\sqrt x + 1}} - \frac{3}{{\sqrt x + 1}} = 1 - \frac{3}{{\sqrt x + 1}}\)

Để P ∈ Z \( \Rightarrow \frac{3}{{\sqrt x + 1}} \in Z \Rightarrow \sqrt x + 1 \in U\left( 3 \right) = \left\{ { \pm 1; \pm 3} \right\}\)

\( + )\sqrt x + 1 = 1 \Leftrightarrow \sqrt x = 0 \Leftrightarrow x = 0\left( {TM} \right)\)

\( + )\sqrt x + 1 = - 1 \Leftrightarrow \sqrt x = - 2(L)\)

\( + )\sqrt x + 1 = 3 \Leftrightarrow \sqrt x  = 2 \Leftrightarrow x = 4\left( L \right)\)

\( + )\sqrt x + 1 = - 3 \Leftrightarrow \sqrt x = - 4\left( L \right)\)

Vậy x = 0 thì P ∈ ℤ.