a) Cho biểu thức A = ( a căn bậc hai a -1 / a - căn bậc hai a - a căn bậc hai a + 1/ a +căn bậc hai a)
a) Với: \[\left\{ \begin{array}{l}a > 0\\a\, \ne \,\left\{ {1\,,\,2} \right\}\end{array} \right.\]
Ta có:
\[A = \left( {\frac{{a\sqrt a - 1}}{{a - \sqrt a }} - \frac{{a\sqrt a + 1}}{{a + \sqrt a }}} \right)\,\,:\,\,\left( {\frac{{a + 2}}{{a - 2}}} \right)\, = \left( {\frac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}} - \frac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right)\,\,:\,\,\left( {\frac{{a + 2}}{{a - 2}}} \right)\]
\[A = \left( {\frac{{a + \sqrt a + 1}}{{\sqrt a }} - \frac{{a - \sqrt a + 1}}{{\sqrt a }}} \right)\,\,:\,\,\left( {\frac{{a + 2}}{{a - 2}}} \right)\, = 2 \cdot \,\left( {\frac{{a - 2}}{{a + 2}}} \right)\, = \frac{{2a - 4}}{{a + 2}}\,\, = \,2 - \frac{8}{{a + 2}}\]
Để \[A \in \mathbb{Z}\,\, \Rightarrow \,\,2 - \frac{8}{{a + 2}}\, \in \,\mathbb{Z}\,\, \Rightarrow \,a + 2\, \in \,U\left( 8 \right)\, = \left\{ { \pm \,1\,;\, \pm \,2\,;\, \pm \,4\,;\, \pm \,8} \right\}\]
Do: \[\left\{ \begin{array}{l}a \in {\mathbb{Z}^ + }\\a\,\, \ne \,\,\left\{ {1\,;\,2} \right\}\end{array} \right.\,\, \Rightarrow \,\,a + 2\,\,\, \ge \,\,5\,\, \Rightarrow \,\,a + \,2\,\, \in \,\,\left\{ {8\,} \right\}\,\, \Rightarrow \,\,a = \,6\,\,\left( {TM} \right)\]
Vậy \[a = \,6\,\,\, \Rightarrow \,A\,\, \in \,\mathbb{Z}\]
b) Đặt :\[M = {x^5} - 2{x^4} - 2021{x^3} + 3{x^2} + 2018x - 2021\]
\[ = {x^5} - 2{x^4} - 2020{x^3} - {x^3} + 2{x^2} + 2020x + {x^2} - 2x - 2020 - 1.\]
\[\begin{array}{l} = {x^3}\left( {{x^2} - 2{x^{}} - 2020} \right) - x\left( {{x^2} - 2{x^{}} - 2020} \right) + \left( {{x^2} - 2x - 2020} \right) - 1\\ = \,\left( {{x^2} - 2x - 2020} \right)\left( {{x^3} - x + 1} \right) - 1\end{array}\]
Mà: \[x = 1 + \sqrt {2021} \,\, \Leftrightarrow \,x - 1 = \sqrt {2021} \,\, \Leftrightarrow \,{\left( {x - 1} \right)^2} = 2021\,\, \Leftrightarrow \,{x^2} - 2x - 2020 = 0.\]
\[ \Rightarrow M = - 1\]