a/ \(A=\sqrt{4+2\sqrt{3}}-\sqrt{5+2\sqrt{6}}+\sqrt{2}\)
\(A=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{2}\)
\(A=\sqrt{3}+1-\sqrt{2}-\sqrt{3}+\sqrt{2}\)
\(A=1\)
b/ \(B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(B=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(B=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(B=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(B=\sqrt{25}=5\)
a)Sửa đề : \(\sqrt{4+2\sqrt{3}}-\sqrt{7+2\sqrt{6}}+\sqrt{2}\)
= \(\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}-\sqrt{\left(\sqrt{6}\right)^2+2\sqrt{6}+1}+\sqrt{2}\)
= \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{6}+1\right)^2}+\sqrt{2}\)
= \(\sqrt{3}+1-\sqrt{6}-1+\sqrt{2}\)
= - \(\sqrt{1}\)
b) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
= \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}\right)^2+4\sqrt{3}+4}}}\)
= \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+2\right)^2}}}\)
= \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}\)
= \(\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
= \(\sqrt{5\sqrt{3}+5\sqrt{\left(\sqrt{3}\right)^2-10\sqrt{3}+25}}\)
= \(\sqrt{5\sqrt{3}+5\sqrt{\left(\sqrt{3}-5\right)^2}}\)
= \(\sqrt{5\sqrt{3}-5\sqrt{3}+5}\)
= \(\sqrt{5}\)
