Cho biểu thức (P) : y = -x^2 và đường thằng (d): y = x-2

a) ĐK: \[\left\{ \begin{array}{l}a \ge 0\\a - 4 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a \ge 0\\a \ne 4\end{array} \right.\].

              b) \[P = \frac{{\sqrt a  + 3}}{{\sqrt a  - 2}} + \frac{{1 - \sqrt a }}{{\sqrt a  + 2}} + \frac{{4 - 4\sqrt a }}{{a - 4}}\]

                       \[ = \frac{{\left( {\sqrt a  + 3} \right).\left( {\sqrt a  + 2} \right) + \left( {1 - \sqrt a } \right).\left( {\sqrt a  - 2} \right) + \left( {4 - 4\sqrt a } \right)}}{{\left( {\sqrt a  - 2} \right).\left( {\sqrt a  + 2} \right)}}\]

                       \[ = \frac{{a + 2\sqrt a  + 3\sqrt a  + 6 + \sqrt a  - 2 - a + 2\sqrt a  + 4 - 4\sqrt a }}{{\left( {\sqrt a  - 2} \right).\left( {\sqrt a  + 2} \right)}}\]

                       \[ = \frac{{4\sqrt a  + 8}}{{\left( {\sqrt a  - 2} \right).\left( {\sqrt a  + 2} \right)}}\]

                       \[ = \frac{{4(\sqrt a  + 2)}}{{\left( {\sqrt a  - 2} \right).\left( {\sqrt a  + 2} \right)}}\]

                       \[ = \frac{4}{{\sqrt a  - 2}}\].

              Vậy \[P = \frac{4}{{\sqrt a  - 2}}\].