Đáp án: A > 5
Giải thích các bước giải:
Ta có: \( A=\frac{3-\sqrt{6+\sqrt{6+\sqrt{6}}}}{3-\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}\)
\( =\frac{\left(3-\sqrt{6+\sqrt{6+\sqrt{6}}}\right).\left(3+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}\right)}{\left(3-\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}\right). \left(3+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}\right)}\)
\( =\frac{\left(3-\sqrt{6+\sqrt{6+\sqrt{6}}}\right).\left(3+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}\right)}{9-6-\sqrt{6+\sqrt{6+\sqrt{6}}}}\)
\(=3+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}\)
Xét \(B=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}\)
\(\Rightarrow B^2=6+\sqrt{6+\sqrt{6+\sqrt{6}}}>2^2=4\)
Suy ra: \(3+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}>3+2=5\)
Vậy \(A>5\)
