Bài 1 : Ta có :
\(A=\sqrt{3x+\sqrt{6x-1}}+\sqrt{3x-\sqrt{6x-1}}\)
\(A\sqrt{2}=\sqrt{6x+2\sqrt{6x-1}}+\sqrt{6x-2\sqrt{6x-1}}\)
\(=\sqrt{6x-1+2\sqrt{6x-1}+1}+\sqrt{6x-1-2\sqrt{6x-1}+1}\)
\(=\sqrt{\left(\sqrt{6x-1}+1\right)^2}+\sqrt{\left(\sqrt{6x-1}-1\right)^2}\)
\(=\left|\sqrt{6x-1}+1\right|+\left|\sqrt{6x-1}-1\right|\)
\(=\sqrt{6x-1}+1+\sqrt{6x-1}-1\)
\(=2\sqrt{6x-1}\)
\(\Rightarrow A=\sqrt{2}\left(\sqrt{6x-1}\right)\)
Thay \(x=4+\sqrt{10}\) vào A ta được :
\(A=\sqrt{2}.\sqrt{6\left(4+\sqrt{10}\right)-1}=\sqrt{2}.\sqrt{24+6\sqrt{10}-1}\)
\(=\sqrt{2}.\sqrt{23+6\sqrt{10}}=\sqrt{46+12\sqrt{10}}\)
\(=\sqrt{36+12\sqrt{10}+10}=\sqrt{\left(6+\sqrt{10}\right)^2}=6+\sqrt{10}\)
Vậy \(A=6+\sqrt{10}\) tại \(x=4+\sqrt{10}\)
