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=\(\sqrt{3-\sqrt{5}}\)\(\sqrt{2}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{6-2\sqrt{5}}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{\left(\sqrt{5}+1\right)^2}\)(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=(\(\sqrt{5}+1\))(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=4(\(3+\sqrt{5}\))
=12+4\(\sqrt{5}\)
a, \(\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}\right)^2-2\sqrt{10}+1}=\sqrt{\left(\sqrt{10}+1\right)^2}\)
\(=\left|\sqrt{10}+1\right|=\sqrt{10}+1\)
b, \(\sqrt{27-10\sqrt{2}}=\sqrt{5^2-10\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(5-\sqrt{2}\right)^2}\)
\(=\left|5-\sqrt{2}\right|=5-\sqrt{2}\)
c, \(\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
làm nốt 2 câu cuối nhé, cách làm y trên
d/\(\sqrt{9+4\sqrt{5}}\)
= \(\sqrt{2^2+4\sqrt{5}+\left(\sqrt{5}\right)^2}\)
=\(\sqrt{\left(2+\sqrt{5}\right)^2}\)
= \(\left|2+\sqrt{5}\right|\)
= \(2+\sqrt{5}\)
e/ \(\sqrt{21+4\sqrt{5}}\)
= \(\sqrt{20+4\sqrt{5}+1}\)
=\(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}+1^2}\)
=\(\sqrt{\left(2\sqrt{5}+1\right)^2}\)
= \(\left|2\sqrt{5}+1\right|\)
= \(2\sqrt{5}+1\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+.......+\frac{\sqrt{n}-\sqrt{n-1}}{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n}-1\right)}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+........+\frac{\sqrt{n}-\sqrt{n-1}}{n-\left(n-1\right)}\)
\(=\sqrt{2}-\sqrt{1}+...........+\sqrt{n}-\sqrt{n-1}\)
\(=\sqrt{n}-\sqrt{1}=\sqrt{n}-1\)
bài B tương tự
a,
\(=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
b, \(\sqrt{8-\sqrt{60}}=\sqrt{8-\sqrt{4.15}}=\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{3}\sqrt{5}}=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{3}\sqrt{5}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\)
2 câu cuối tự làm nhé