Đáp án:
2) \(P = 1 - \sqrt 2 \)
Giải thích các bước giải:
\(\begin{array}{l}
1)P = \dfrac{{x + 2\sqrt x - 11 + \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x - 11 + x - 1 - x + 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x - 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
2)Thay:x = 3 - 2\sqrt 2 \\
= 2 - 2\sqrt 2 .1 + 1 = {\left( {\sqrt 2 - 1} \right)^2}\\
\to P = \dfrac{{\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} - 1}}{{\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} + 1}} = \dfrac{{\sqrt 2 - 1 - 1}}{{\sqrt 2 - 1 + 1}} = \dfrac{{\sqrt 2 - 2}}{{\sqrt 2 }}\\
= 1 - \sqrt 2
\end{array}\)
