10.
a) Điều kiện: [imath]x \geq 0, x \neq 1[/imath]
[imath]P=\dfrac{2 \sqrt{x} (1-x)^2}{2+ \sqrt{x}}: \left [ \left ( \dfrac{x \sqrt{x}-1}{\sqrt{x}-1} + \sqrt{x} \right ) \left ( \dfrac{x \sqrt{x}+1}{\sqrt{x}+1} - \sqrt{x} \right ) \right ] \\
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= \dfrac{2 \sqrt{x} [(-1)(x-1)]^2}{2+ \sqrt{x}}: \left \{ \left [ \dfrac{( \sqrt{x}-1)(x+ \sqrt{x}+1)}{\sqrt{x}-1} + \sqrt{x} \right ] \left [ \dfrac{( \sqrt{x}+1)(x- \sqrt{x}+1)}{\sqrt{x}+1} - \sqrt{x} \right ] \right \} \\ \\
= \dfrac{2 \sqrt{x} (x-1)^2}{2+ \sqrt{x}}: \left \{ \left [ \dfrac{( \sqrt{x}-1)(x+ \sqrt{x}+1)}{\sqrt{x}-1} + \sqrt{x} \right ] \left [ \dfrac{( \sqrt{x}+1)(x- \sqrt{x}+1)}{\sqrt{x}+1} - \sqrt{x} \right ] \right \} \\
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= \dfrac{2 \sqrt{x} (x-1)^2}{2+ \sqrt{x}}: [(x+ \sqrt{x}+1+ \sqrt{x}) (x- \sqrt{x}+1 -\sqrt{x})] \\
= \dfrac{2 \sqrt{x} (x-1)^2}{2+ \sqrt{x}}: [(x+ 2 \sqrt{x}+1) (x- 2 \sqrt{x}+1)] \\
= \dfrac{2 \sqrt{x} (x-1)^2}{2+ \sqrt{x}}: [( \sqrt{x}+1)^2 ( \sqrt{x} -1 )^2] \\
= \dfrac{2 \sqrt{x} (x-1)^2}{2+ \sqrt{x}}. \dfrac{1}{( \sqrt{x}+1)^2 ( \sqrt{x} -1 )^2} \\ \\
= \dfrac{2 \sqrt{x} (x-1)^2}{2+ \sqrt{x}}. \dfrac{1}{[( \sqrt{x}+1) ( \sqrt{x} -1 )]^2} \\ \\
= \dfrac{2 \sqrt{x} (x-1)^2}{2+ \sqrt{x}}. \dfrac{1}{(x-1)^2} \\
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= \dfrac{2 \sqrt{x} }{2+ \sqrt{x}}
[/imath]
b) [imath]P = \dfrac{2 \sqrt{x} }{2+ \sqrt{x}} = \dfrac{2 \sqrt{x} +4-4}{2+ \sqrt{x}} \\
= \dfrac{2 ( \sqrt{x} +2)-4}{2+ \sqrt{x}} \\
=2- \dfrac{4}{2+ \sqrt{x}}[/imath]
[imath]P \in \mathbb{Z} \Leftrightarrow \dfrac{4}{2+ \sqrt{x}} \in \mathbb{Z}[/imath]
[imath](2+ \sqrt{x}) \in Ư(4)= \left \{ \pm 1; \pm 2; \pm 4 \right \}[/imath]
Giải ra và kết hợp đối chiếu điều kiện, ta có: [imath]x=0; x=4[/imath]
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