Đặt \(\left(\sqrt{3}+\sqrt{2}\right)^x=t>0\Rightarrow\left(\sqrt{3}-\sqrt{2}\right)^x=\dfrac{1}{t}\)
BPT trở thành:
\(t+\dfrac{1}{t}\le2\)
\(\Leftrightarrow t^2-2t+1\le0\)
\(\Leftrightarrow\left(t-1\right)^2\le0\)
\(\Rightarrow t-1=0\)
\(\Rightarrow t=1\)
\(\Rightarrow\left(\sqrt{3}+\sqrt{2}\right)^x=1\)
\(\Rightarrow x=0\)