b.
Ta có:
$(\sqrt 3+\sqrt 2)^x+(\sqrt 3-\sqrt 2)^x=10$
$\Leftrightarrow (\sqrt 3+\sqrt 2)^x+\dfrac{1}{(\sqrt 3+\sqrt 2)^x}=10$
$\Leftrightarrow (\sqrt 3+\sqrt 2)^{2x}-10(\sqrt 3+\sqrt 2)^x+1=0$
$\Leftrightarrow \left[\begin{array}{l}(\sqrt 3+\sqrt 2)^x=5+2\sqrt 6\\(\sqrt 3+\sqrt 2)^x=5-2\sqrt 6\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}(\sqrt 3+\sqrt 2)^x=(\sqrt 3+\sqrt 2)^2\\(\sqrt 3+\sqrt 2)^x=(\sqrt 3+\sqrt 2)^{-2}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x=2\\x=-2\end{array}\right.$