b. Ta có:
\(M=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\\ =\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\\ =\dfrac{\sqrt{\left(\sqrt{7}\right)^2+2\cdot\sqrt{7}\cdot1+1^2}-\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot1+1^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{\left(\sqrt{7}+1\right)}^2-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}\\ =\dfrac{2}{\sqrt{2}}=\sqrt{2}\\ N=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\\ =\dfrac{\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\\ =\dfrac{\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}-\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\\ =\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\\ =\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
=> M = N
c) Ta có:
\(\left(C+E\right)^2\\ =\left(\sqrt{45+\sqrt{2009}}+\sqrt{45-\sqrt{2009}}\right)^2\\ =\left(\sqrt{45+\sqrt{2009}}\right)^2+2\sqrt{45+\sqrt{2009}}\cdot\sqrt{45-\sqrt{2009}}+\left(\sqrt{45-\sqrt{2009}}\right)^2\\ =45+\sqrt{2009}+2\sqrt{45^2-\left(\sqrt{2009}\right)^2}+45-\sqrt{2009}\\ =\\ =90+2\sqrt{16}\\ =90+8\\ =98\\ =>C+E=\sqrt{98}=7\sqrt[]{2}\)
=> Đpcm
