\(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}=\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}=\left|\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right|+\left|\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right|=\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}=2\sqrt{\dfrac{7}{2}}=\sqrt{14}\)
\(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\\ =\dfrac{1}{\sqrt{2}}\left(\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}\right)\\ =\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{7}+1\right)^2}\right)\\ =\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{7}-1\right|+\left|\sqrt{7}+1\right|\right)\\ =\dfrac{1}{\sqrt{2}}\left(\sqrt{7}-1+\sqrt{7}+1\right)\\ =\dfrac{1}{\sqrt{2}}.2\sqrt{7}=\sqrt{14}\)
đặt biểu thức trên là A ta có
A=√4−√7+√4+√7
\(A^2=\left(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\right)^2\)
\(A^2=4-\sqrt{7}+2.\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\)
\(A^2=8+2.\sqrt{16-7}\)
A2=8+2.3
A2=14
A=\(\sqrt{14}\)
\(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\)
\(=\dfrac{\sqrt{7}-1+\sqrt{7}+1}{\sqrt{2}}\)
\(=\sqrt{14}\)
