\(\left(\frac{\sqrt{8}}{3}\right)^2=\frac{8}{9}>\frac{9}{16}=\left(\frac{3}{4}\right)^2\Rightarrow\frac{\sqrt{8}}{3}>\frac{3}{4}\)

a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)

=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)

=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)

=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)

=-7

b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)

So sánh:

1) \(2\sqrt{27}\) và \(\sqrt{147}\)

+ \(2\sqrt{27}\) = \(6\sqrt{3}\)

+ \(\sqrt{147}\) = \(7\sqrt{3}\)

⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)

Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)

2) \(2\sqrt{15}\) và \(\sqrt{59}\)

+ \(2\sqrt{15}\) = \(\sqrt{60}\)

⇒ \(\sqrt{60}\) > \(\sqrt{59}\)

Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)

3) \(2\sqrt{2}-1\) và 2

\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)

So sánh: 3 và \(2\sqrt{2}\)

+ 3 = \(\sqrt{9}\)

+ \(2\sqrt{2}=\sqrt{8}\)

⇒ \(\sqrt{8}\) < \(\sqrt{9}\)

⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1

⇒ \(2\sqrt{2}\) - 1 < 3 - 1

Vậy: \(2\sqrt{2}-1< 2\)

4) \(\frac{\sqrt{3}}{2}\) và 1

+ 1 = \(\frac{2}{2}\)

⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)

Vậy: \(\frac{\sqrt{3}}{2}\) < 1

5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)

+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)

⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)

Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)

Bạn xem hộ mk đề cậu b nhé căn 5- căn 2 hay là căn 5 - 2

a/ \(\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8\left(1+\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)

b/ \(\frac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}=\frac{-2}{\sqrt{6}}-\frac{1}{\sqrt{6}}=\frac{-3}{\sqrt{6}}=-\frac{\sqrt{6}}{2}\)

c/ \(\frac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}+\frac{\sqrt{\left(2+\sqrt{3}\right)^2}}{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}=2-\sqrt{3}+2+\sqrt{3}=4\)

d/ \(\frac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\frac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

\(=\frac{\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)}{8}=\frac{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}=1\)

e/ \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{\sqrt{2}}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}\)

\(=\frac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\frac{\sqrt{2}}{3+\sqrt{3}}+\frac{\sqrt{2}}{3-\sqrt{3}}=\frac{\sqrt{2}\left(3-\sqrt{3}+3+\sqrt{3}\right)}{6}=\sqrt{2}\)

f/ \(\frac{9+4\sqrt{5}-8\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-2}{2}\)

a/\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=2\sqrt{5}+\frac{8}{1-\sqrt{5}}\)

\(=\frac{2\sqrt{5}-10+8}{1-\sqrt{5}}=\frac{-2\left(1-\sqrt{5}\right)}{1-\sqrt{5}}=-2\)

b/Đề sai

c/\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\frac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{\sqrt{2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

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

d/ \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}=\frac{9+4\sqrt{5}-8\sqrt{5}}{2\sqrt{5}-4}=\frac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\frac{\left(\sqrt{5}-2\right)^2}{2\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-2}{2}\)

a: \(=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}\)

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

b: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)

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

d: \(=-\left(\sqrt{5}+\sqrt{2}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)=-3\)