Đáp án:
2) \(P = \dfrac{{ - 1 - 2\sqrt 2 }}{7}\)
Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
a)P = \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right) - 3\left( {\sqrt x - 1} \right) + 3}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - 4 - 3\sqrt x + 3 + 3}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - 3\sqrt x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 2}}\\
2)Thay:x = 6 - 4\sqrt 2 = 4 - 2.2.\sqrt 2 + 2\\
= {\left( {2 - \sqrt 2 } \right)^2}\\
\to P = \dfrac{{\sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} - 2}}{{\sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} + 2}} = \dfrac{{2 - \sqrt 2 - 2}}{{2 - \sqrt 2 + 2}} = \dfrac{{ - 1 - 2\sqrt 2 }}{7}
\end{array}\)
