Question

Answer 1

Bài 5:

a: \(A=\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}-\sqrt{3}\right)}=\dfrac{\sqrt{5}}{2}\)

b: \(B=\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{2}-\sqrt{5}\right)}{-\sqrt{7}\left(\sqrt{2}-\sqrt{5}\right)}=-\sqrt{\dfrac{3}{7}}=-\dfrac{\sqrt{21}}{7}\)

c: \(C=\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\)

d: \(D=\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

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

\(=\sqrt{5}+\dfrac{\sqrt{5}}{2}=\dfrac{3\sqrt{5}}{2}\)

Bài 6:

a: \(A=\sqrt{32\cdot200}=\sqrt{64\cdot100}=8\cdot10=80\)

b: \(B=\sqrt{5}\cdot\sqrt{125}=\sqrt{5}\cdot5\sqrt{5}=25\)

c: \(C=\sqrt{\left(9+2\sqrt{14}\right)\left(9-2\sqrt{14}\right)}\)

\(=\sqrt{81-56}\)

\(=\sqrt{25}=5\)

d: \(D=\sqrt{117,5^2-26,5^2-1440}\)

\(=\sqrt{\left(117,5-26,5\right)\left(117,5+26,5\right)-1440}\)

\(=\sqrt{91\cdot144-1440}=\sqrt{144\cdot81}=12\cdot9=108\)

Answer 2

Bài 6: 
a, A= \(\sqrt{32.200}=\sqrt{32.2.100}=\sqrt{64.100}=\sqrt{8^2.10^2}=8.10=80\)
b, B\(=\sqrt{5}.\sqrt{125}=\sqrt{5}.\sqrt{5^3}=\sqrt{5^4}=25\)
c, C=\(\sqrt{\left(9+2\sqrt{14}\right).\left(9-2\sqrt{14}\right)}=\sqrt{9^2-\left(2\sqrt{14}\right)^2}=\sqrt{25}=5\)
d, D = \(\sqrt{117,5^2-26,5^2-1440}=\sqrt{\left(117,5-26,5\right).\left(117,5+26,5\right)-1440}=\sqrt{13104-1440}=\sqrt{11664}=108\)

Answer 3

a,\(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right).\sqrt{6}\)
\(=\sqrt{\dfrac{2}{3}.6}+\sqrt{\dfrac{50}{3}.6}-\sqrt{24.6}\)
\(=\sqrt{4}+\sqrt{100}-\sqrt{144}\)
= 2 + 10 -12 =0
b, \(\sqrt{3+\sqrt{5}}.\sqrt{2}=\sqrt{\left(3+\sqrt{5}\right).2}=\sqrt{6+2\sqrt{5}}=\sqrt{5+2.\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
c, (\(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\sqrt{\dfrac{4}{3}}\)) .\(\sqrt{12}=\sqrt{\dfrac{3}{4}.12}-\sqrt{3.12}+5\sqrt{\dfrac{4}{3}.12}=\sqrt{9}-\sqrt{36}+5\sqrt{16}=3-6+5.4=17\)
d,