\(\sqrt{\left|4\sqrt{6}-11\right|}-\sqrt{4\sqrt{6}+11}\)
Vì \(4\sqrt{6}< 11\) nên khi thoát dấu GTTĐ, ta được:
\(\sqrt{11-4\sqrt{6}}-\sqrt{11+4\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\left(2\sqrt{2}\right).\sqrt{3}+\left(2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2+2.\left(2\sqrt{2}\right).\sqrt{3}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(\sqrt{3}-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}+2\sqrt{2}\right)^2}\)
=|√3-2√2|-|√3+2√2|
= 2√2-√3-√3-2√2
= -2√3
\(\sqrt{\left|4\sqrt{6}-11\right|}-\sqrt{4\sqrt{6}+11}\)
Ta có:
\(4\sqrt{6}< 11\)
\(\Rightarrow\sqrt{11-4\sqrt{6}}-\sqrt{11+4\sqrt{6}}\)
\(\Rightarrow\sqrt{\left(\sqrt{3}\right)^2-2\left(2\sqrt{2}\right)\sqrt{3}+\left(2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2+2\left(2\sqrt{2}\right)\sqrt{3}+\left(2\sqrt{2}\right)^2}\)
Từ đây rút gọn căn của 2 bên rồi tính nốt