Đáp án:
\(\begin{array}{l}
a)\quad P =-12 + 3\sqrt2\\
b)\quad P = 1\\
c)\quad P = 4\sqrt5\\
d)\quad P = \sqrt7
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\quad P = \left(2 + 3\sqrt2\right)\left(1 - \sqrt{2\left(3 - \sqrt2\right)^2}\right)\\
\to P =\left(2 + 3\sqrt2\right)\left(1 - \sqrt2\left|3 - \sqrt2\right|\right)\\
\to P = \left(2 + 3\sqrt2\right)\left(3 - 3\sqrt2\right)\\
\to P =-12 + 3\sqrt2\\
b)\quad P = P = \left(\dfrac{3 + \sqrt6}{\sqrt8 + \sqrt{12}} + \dfrac{\sqrt{27}}{6} \right)\cdot \dfrac{1}{\sqrt3}\\
\to P = \left[\dfrac{\sqrt3\left(\sqrt3 + \sqrt2\right)}{2\left(\sqrt2 + \sqrt{3}\right)} + \dfrac{3\sqrt{3}}{6} \right]\cdot \dfrac{1}{\sqrt3}\\
\to P = \left(\dfrac{\sqrt3}{2} + \dfrac{\sqrt3}{2}\right)\cdot \dfrac{1}{\sqrt3}\\
\to P = \sqrt3\cdot \dfrac{1}{\sqrt3}\\
\to P = 1\\
c)\quad P = \dfrac{\sqrt5 + 5}{\sqrt5 + 1} + \dfrac{\sqrt{15} - 5}{\sqrt3 - 1} + \dfrac{\sqrt{60} +10}{\sqrt3 + \sqrt5}\\
\to P = \dfrac{\sqrt5\left(1 + \sqrt5\right)}{\sqrt5 + 1} + \dfrac{\sqrt5\left(\sqrt3 - 1\right)}{\sqrt3- 1} + \dfrac{2\sqrt5\left(\sqrt3 + \sqrt5\right)}{\sqrt3 + \sqrt5}\\
\to P = \sqrt5 + \sqrt5 + 2\sqrt5\\
\to P = 4\sqrt5\\
d)\quad P = \sqrt{3 +\sqrt2}\ \sqrt{3 + \sqrt{6+\sqrt2}}\ \sqrt{3 + \sqrt{6 + \sqrt{6+\sqrt2}}}\ \sqrt{3 - \sqrt{6 + \sqrt{6+\sqrt2}}}\\
\to P = \sqrt{3 +\sqrt2}\ \sqrt{3 + \sqrt{6+\sqrt2}}\ \sqrt{\left(3 + \sqrt{6 + \sqrt{6+\sqrt2}}\right)\left(3 - \sqrt{6 + \sqrt{6+\sqrt2}}\right)}\\
\to P = \sqrt{3 +\sqrt2}\ \sqrt{3 + \sqrt{6+\sqrt2}}\ \sqrt{9 - \left(6 + \sqrt{6+\sqrt2} \right)}\\
\to P = \sqrt{3 +\sqrt2}\ \sqrt{3 + \sqrt{6+\sqrt2}}\ \sqrt{3 - \sqrt{6+\sqrt2}}\\
\to P = \sqrt{3 +\sqrt2}\ \sqrt{\left(3 + \sqrt{6+\sqrt2}\right)\left(3 - \sqrt{6+\sqrt2}\right)}\\
\to P = \sqrt{3 +\sqrt2}\ \sqrt{9 - \left(6 + \sqrt2\right)}\\
\to P = \sqrt{3 +\sqrt2}\ \sqrt{3 - \sqrt2}\\
\to P = \sqrt{\left(3 + \sqrt2\right)\left(3 - \sqrt2\right)}\\
\to P = \sqrt{9 - 2}\\
\to P = \sqrt7
\end{array}\)
