Question

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Answer 1

$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
1)\\
A = \dfrac{{3\left( {x + \sqrt x - 3} \right)}}{{x + \sqrt x - 2}} + \dfrac{{\sqrt x + 3}}{{\sqrt x + 2}} - \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}\\
= \dfrac{{3x + 3\sqrt x - 9}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}} + \dfrac{{\sqrt x + 3}}{{\sqrt x + 2}} - \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}\\
= \dfrac{{3x + 3\sqrt x - 9 + \left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x + 3\sqrt x - 9 + x + 2\sqrt x - 3 - \left( {x - 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{4x + 5\sqrt x - 12 - x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x + 5\sqrt x - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x + 8\sqrt x - 3\sqrt x - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {3\sqrt x + 8} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3\sqrt x + 8}}{{\sqrt x + 2}}\\
2)A = \dfrac{{3\sqrt x + 8}}{{\sqrt x + 2}}\\
= \dfrac{{3\sqrt x + 6 + 2}}{{\sqrt x + 2}}\\
= \dfrac{{3.\left( {\sqrt x + 2} \right) + 2}}{{\sqrt x + 2}}\\
= 3 + \dfrac{2}{{\sqrt x + 2}}\\
A \in Z\\
\Leftrightarrow \dfrac{2}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \sqrt x + 2 = 2\left( {do:\sqrt x + 2 \ge 2} \right)\\
\Leftrightarrow \sqrt x = 0\\
\Leftrightarrow x = 0\left( {tm} \right)
\end{array}$

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Answer 2

$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
1)\\
A = \dfrac{{3\left( {x + \sqrt x - 3} \right)}}{{x + \sqrt x - 2}} + \dfrac{{\sqrt x + 3}}{{\sqrt x + 2}} - \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}\\
= \dfrac{{3x + 3\sqrt x - 9}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}} + \dfrac{{\sqrt x + 3}}{{\sqrt x + 2}} - \dfrac{{\sqrt x - 2}}{{\sqrt x - 1}}\\
= \dfrac{{3x + 3\sqrt x - 9 + \left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x + 3\sqrt x - 9 + x + 2\sqrt x - 3 - \left( {x - 4} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{4x + 5\sqrt x - 12 - x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x + 5\sqrt x - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3x + 8\sqrt x - 3\sqrt x - 8}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {3\sqrt x + 8} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{3\sqrt x + 8}}{{\sqrt x + 2}}\\
2)A = \dfrac{{3\sqrt x + 8}}{{\sqrt x + 2}}\\
= \dfrac{{3\sqrt x + 6 + 2}}{{\sqrt x + 2}}\\
= \dfrac{{3.\left( {\sqrt x + 2} \right) + 2}}{{\sqrt x + 2}}\\
= 3 + \dfrac{2}{{\sqrt x + 2}}\\
A \in Z\\
\Leftrightarrow \dfrac{2}{{\sqrt x + 2}} \in Z\\
\Leftrightarrow \sqrt x + 2 = 2\left( {do:\sqrt x + 2 \ge 2} \right)\\
\Leftrightarrow \sqrt x = 0\\
\Leftrightarrow x = 0\left( {tm} \right)
\end{array}$

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