Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
C3:\\
a.ĐK:a \ge 0;a \ne 1\\
b.P = \frac{{2{a^2} + 4 - \left( {1 - \sqrt a } \right)\left( {1 + a + {a^2}} \right) - \left( {1 + \sqrt a } \right)\left( {1 + a + {a^2}} \right)}}{{\left( {1 - a} \right)\left( {1 + a + {a^2}} \right)}}\\
= \frac{{2{a^2} + 4 - 1 - a - {a^2} + \sqrt a + a\sqrt a + {a^2}\sqrt a - 1 - a - {a^2} - \sqrt a - a\sqrt a - {a^2}\sqrt a }}{{\left( {1 - a} \right)\left( {1 + a + {a^2}} \right)}}\\
= \frac{{2 - 2a}}{{\left( {1 - a} \right)\left( {1 + a + {a^2}} \right)}} = \frac{2}{{1 + a + {a^2}}}
\end{array}\)
\(\begin{array}{l}
C4:\\
a.ĐK:x \ge 0;x \ne 3\\
b.A = \frac{{x - 2\sqrt 3 .\sqrt x + 3}}{{x - 3}}.\sqrt 4 \left( {\sqrt x + \sqrt 3 } \right)\\
= \frac{{{{\left( {\sqrt x - \sqrt 3 } \right)}^2}}}{{\left( {\sqrt x - \sqrt 3 } \right)\left( {\sqrt x + \sqrt 3 } \right)}}.\sqrt 4 \left( {\sqrt x + \sqrt 3 } \right)\\
= 2\left( {\sqrt x - \sqrt 3 } \right)\\
c.x = 4 - 2\sqrt 3 \\
\to A = 2\left( {\sqrt {4 - 2\sqrt 3 } - \sqrt 3 } \right) = - 2
\end{array}\)
