Question

[Võ Thiên Long]

Câu 1:Rút gọn biểu thức

a.A =  \(\frac{3-\sqrt{3}}{\sqrt{3}}\)  b.B  =  \(\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}\)c. C  =  \(\frac{2\sqrt{2}+\sqrt{6}}{4+\sqrt{12}}\) d. D = \(\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}\)

e.E =\(\frac{1-\sqrt{a^3}}{a-1}\)  f. F = \(\frac{\sqrt{7-4\sqrt{3}}}{\sqrt{3}-2}\)   g. G  = \(3.\sqrt{\frac{12\left(a-2\right)^2}{27}}\) h. H = \(\left(a-b\right).\sqrt{\frac{ab}{\left(a-b\right)^2}}\)

Answer 1

[KAl(SO4)2·12H2O]

a) \(A=\frac{3-\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}}=\sqrt{3}-1\)

b) \(B=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{5}+1}=\frac{\sqrt{5}+1}{\sqrt{5}+1}=1\)

c) \(C=\frac{2\sqrt{2}+\sqrt{6}}{4+\sqrt{12}}=\frac{2\sqrt{2}+\sqrt{6}}{4+2\sqrt{3}}=\frac{\left(2\sqrt{2}+\sqrt{6}\right)\left(4-2\sqrt{3}\right)}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}\)

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