Toán sao đưa tin vào đây
Đặt \(B=\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)
\(\Rightarrow B^2=\left(\sqrt{4-\sqrt{15}}\right)^2-2.\sqrt{4-\sqrt{15}}.\sqrt{4+\sqrt{15}}+\left(\sqrt{4+\sqrt{15}}\right)^2\)
\(\Rightarrow B^2=4-\sqrt{15}-2.\sqrt{4^2-\left(\sqrt{15}\right)^2}+4+\sqrt{15}\)
\(\Rightarrow B^2=8-2.1=6\Rightarrow\left[{}\begin{matrix}B=\sqrt{6}\\B=-\sqrt{6}\end{matrix}\right.\)
Vì \(\sqrt{4-\sqrt{15}}< \sqrt{4+\sqrt{15}}\) nên B<0 \(\Rightarrow B=-\sqrt{6}\)
\(\Rightarrow\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}=-\sqrt{6}+\sqrt{6}=0\)
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