Giải
4 + \(\sqrt{7}\) = \(\dfrac{8+2\sqrt{7}}{2}\) = \(\dfrac{7+2\sqrt{7}+1}{2}\) = \(\left(\dfrac{\sqrt{7}+1}{\sqrt{2}}\right)^2\)
4 - \(\sqrt{7}\) = \(\dfrac{8-2\sqrt{7}}{2}\) = \(\dfrac{7-2\sqrt{7}+1}{2}\) = \(\left(\dfrac{\sqrt{7}-1}{\sqrt{2}}\right)^2\)
Do đó M = \(\sqrt{\left(\dfrac{\sqrt{7+1}}{\sqrt{2}}\right)^2}-\sqrt{\left(\dfrac{\sqrt{7}-1}{\sqrt{2}}\right)^2}=\dfrac{\sqrt{7}+1}{\sqrt{2}}-\dfrac{\sqrt{7}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(M=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{\left(\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\dfrac{\sqrt{7}}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{\sqrt{7}}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}=\sqrt{2}\)
\(M=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\\ M^2=4+\sqrt{7}+4-\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}\\ M^2=8-2\sqrt{16-7}=8-2\sqrt{9}=2\\ \Rightarrow M=\sqrt{2}\)
rút gọn M= \(\sqrt{4+\sqrt{ }7}\)- \(\sqrt{4-\sqrt{ }7}\)
M= 2+1,6 - 2-1,6
M= 0
