Cho M = 2 / (căn bậc hai x - 1) + 2 / (căn bậc hai x + 1) - (5 - căn bậc hai x)

a) \[M = \frac{2}{{\sqrt x - 1}} + \frac{2}{{\sqrt x + 1}} - \frac{{5 - \sqrt x }}{{x - 1}}\]

\[ = \frac{{2\left( {\sqrt x + 1} \right) + 2\left( {\sqrt x - 1} \right) - 5 + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\]

\[ = \frac{{2\sqrt x + 2 + 2\sqrt x - 2 - 5 + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\]

\[ = \frac{{5\sqrt x - 5}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\]

\[ = \frac{{5\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\]

\[ = \frac{5}{{\sqrt x + 1}}\]

b) Với x = 4 thì \[M = \frac{5}{{\sqrt x + 1}} = \frac{5}{{\sqrt 4 + 1}} = \frac{5}{3}\]

c) M có giá trị nguyên \[ \Leftrightarrow \frac{5}{{\sqrt x + 1}}\] có giá trị nguyên.

\[ \Leftrightarrow 5 \vdots \sqrt x + 1\]\[ \Rightarrow \sqrt x + 1 \in \left\{ {1;5} \right\}\]

\[\sqrt x + 1 = 1 \Rightarrow x = 0\]

\[\sqrt x + 1 = 5 \Rightarrow x = 16\]