Biến đổi về dạng hằng đẳng thức tính giá trị biểu thức.
a, A = $\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}$
= $\sqrt{2+2\sqrt{2.5}+5}-\sqrt{5-2\sqrt{5.2}+2}$
= $\sqrt{(\sqrt{2})^{2}+2\sqrt{2}.\sqrt{5}+(\sqrt{5})^{2}}-\sqrt{(\sqrt{5})^{2}-2\sqrt{2}.\sqrt{5}+(\sqrt{2})^{2}}$
= $\sqrt{(\sqrt{2}+\sqrt{5})^{2}}-\sqrt{(\sqrt{5}-\sqrt{2})^{2}}$
= $|\sqrt{2}+\sqrt{5}|-|\sqrt{5}-\sqrt{2}|$ = $\sqrt{2}+\sqrt{5}-\sqrt{5}+\sqrt{2}$ = $2\sqrt{2}$
b, B = $\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{3-2\sqrt{2}}}}$
= $\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{2-2\sqrt{2}+1}}}$ = $\sqrt{4\sqrt{2}+4\sqrt{10-8\sqrt{(\sqrt{2}-1)^{2}}}}$
= $\sqrt{4\sqrt{2}+4\sqrt{18-8\sqrt{2}}}$ = $\sqrt{4\sqrt{2}+4\sqrt{4^{2}-2.4.\sqrt{2}+2}}$
= $\sqrt{4\sqrt{2}+4\sqrt{(4-\sqrt{2})^{2}}}$ = $\sqrt{4\sqrt{2}+4.(4-\sqrt{2})}$
= $\sqrt{4\sqrt{2}+4.(4-\sqrt{2})}$ = $\sqrt{16}$ = 4
c, C = $\sqrt{3+\sqrt{13+\sqrt{48}}}$ = $\sqrt{3+\sqrt{13+\sqrt{16.3}}}$
= $\sqrt{3+\sqrt{13+\sqrt{16}.\sqrt{3}}}$ = $\sqrt{3+\sqrt{13+4.\sqrt{3}}}$
= $\sqrt{3+\sqrt{(2\sqrt{3})^{2}+2.2.\sqrt{3}+1}}$ = $\sqrt{3+\sqrt{(2\sqrt{3}+1)^{2}}}$
= $\sqrt{3+2\sqrt{3}+1}$ = $\sqrt{(\sqrt{3}+1)^{2}}$ = $\sqrt{3}+1$