Question

Tính: a. \(\left(\sqrt{10}+\sqrt{2}\right)\cdot\left(6-2\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}\)

b. \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)

c. \(\sqrt{3,5-\sqrt{6}}+\sqrt{3,5+\sqrt{6}}\)

d, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)

e, \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)

Answer 1

e) \(E=A-\sqrt{2}\)

\(A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

\(A^2=8-2\sqrt{16-7}=8-6=2\)

\(A>0=>A=\sqrt{2}\)

\(E=A-\sqrt{2}=0\)

Answer 2

a)\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)

=\(\left(6\sqrt{10}+6\sqrt{2}-10\sqrt{2}-2\sqrt{10}\right)\sqrt{3+\sqrt{5}}\)

=\(\left(4\sqrt{10}-4\sqrt{2}\right)\sqrt{3+\sqrt{5}}=\left(4\sqrt{10}-4\sqrt{2}\right)\dfrac{\sqrt{5}+1}{2}\)

=\(\dfrac{20\sqrt{2}+4\sqrt{10}-4\sqrt{10}-4\sqrt{2}}{2}\)

=\(\dfrac{16\sqrt{2}}{2}=8\sqrt{2}\)

b)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)

=\(\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}=0\)

c)\(\sqrt{3,5-\sqrt{6}}+\sqrt{3,5+\sqrt{6}}\)

=\(\dfrac{\sqrt{6}-1+\sqrt{6}+1}{\sqrt{2}}=2\sqrt{3}\)

d)\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)

=\(\dfrac{\sqrt{7}-1-\sqrt{7}-1+\sqrt{14}}{\sqrt{2}}=\sqrt{7}-1\)

e)\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)

=\(\dfrac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}=0\)

Answer 3

a) \(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right).\sqrt{3+\sqrt{5}}\)

\(=\sqrt{2}\left(\sqrt{5}+1\right).2.\left(3-\sqrt{5}\right).\sqrt{3+\sqrt{5}}\)

\(=\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right).\sqrt{6+2\sqrt{5}}\)

\(=\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right).\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right).\left(\sqrt{5}+1\right)\)

\(=\left(\sqrt{5}+1\right)^2.\left(3-\sqrt{5}\right)\)

\(=\left(6+2\sqrt{5}\right)\left(3-\sqrt{5}\right)\)

\(=18-6\sqrt{5}+6\sqrt{5}-10\)

\(=8\)

Answer 4

b) \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)

\(=\dfrac{\sqrt{2}.\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\right)}{\sqrt{2}}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2}{\sqrt{2}}\)

\(=\dfrac{\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+1-\sqrt{5}+1-2}{\sqrt{2}}\)

\(=\dfrac{0}{\sqrt{2}}=0\)

Answer 5

c) \(\sqrt{3,5-\sqrt{6}}+\sqrt{3,5+\sqrt{6}}\)

\(=\sqrt{\dfrac{7}{2}-\sqrt{6}}+\sqrt{\dfrac{7}{2}+\sqrt{6}}\)

\(=\sqrt{7-2\sqrt{6}}+\sqrt{7+2\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}+1\right)^2}\)

\(=\left|\sqrt{6}-1\right|+\left|\sqrt{6}+1\right|\)

\(=\sqrt{6}-1+\sqrt{6}+1\) ( vì \(\sqrt{6}-1>0;\sqrt{6}+1>0\) )

\(=2\sqrt{6}\)

Answer 6

d) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)\(=\dfrac{\sqrt{2}.\left(\sqrt{4-\sqrt{7}}-\sqrt{4}+\sqrt{7}+\sqrt{7}\right)}{\sqrt{2}}\)

\(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+2\sqrt{7}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+2\sqrt{7}}{\sqrt{2}}\)

\(=\dfrac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|+2\sqrt{7}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1+2\sqrt{7}}{\sqrt{2}}\)

\(=\dfrac{2\sqrt{7}-2}{\sqrt{2}}\)

\(=\dfrac{2\left(\sqrt{7}-1\right)}{\sqrt{2}}\)

\(=\sqrt{2}\left(\sqrt{7}-1\right)\)

Answer 7

e) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)

\(=\dfrac{\sqrt{2}\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\right)}{\sqrt{2}}\)

\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}-2}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}\)

\(=\dfrac{0}{\sqrt{2}}=0\)