a) 3\(\sqrt{3}\)=\(\sqrt{27}\)>\(\sqrt{12}\)

c) \(\frac{1}{3}\)\(\sqrt{51}\)=\(\sqrt{\frac{51}{9}}\)<\(\frac{1}{5}\)\(\sqrt{150}\)=\(\sqrt{\frac{150}{25}}\)=\(\sqrt{6}\)

b) 3\(\sqrt{5}\)=\(\sqrt{45}\)< 7=\(\sqrt{49}\)

d) \(\frac{1}{2}\sqrt{6}\)=\(\sqrt{\frac{6}{4}}\)=\(\sqrt{\frac{3}{2}}\)< 6\(\sqrt{\frac{1}{2}}\)=\(\sqrt{\frac{36}{2}}\)=\(\sqrt{18}\)

a) Ta có: $3\sqrt{3}=\sqrt{3^2.3}=\sqrt{9.3}=\sqrt{27}$

Vì $\sqrt{27}>\sqrt{12}$ nên $3\sqrt{3}>\sqrt{12}$

Vậy $3\sqrt{3}>\sqrt{12}$.
b) Ta có: $3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}$

$7=\sqrt{7^2}=\sqrt{49}$

Vì $\sqrt{49}>\sqrt{45}$ nên $7>3\sqrt{5}$

Vậy $7>3\sqrt{5}$.

c) Ta có: $\frac{1}{3}\sqrt{51}=\sqrt{{\left(\frac{1}{3}\right)}^2.51}=\sqrt{\frac{51}{9}}$

$\frac{1}{5}\sqrt{150}=\sqrt{{\left(\frac{1}{5}\right)}^2.150}=\sqrt{\frac{150}{25}}=\sqrt{6}=\sqrt{\frac{6.9}{9}}=\sqrt{\frac{54}{9}}$

Vì $\sqrt{\frac{54}{9}}>\sqrt{\frac{51}{9}}$ nên $\frac{1}{3}\sqrt{51}<\frac{1}{5}\sqrt{150}$

Vậy $\frac{1}{3}\sqrt{51}<\frac{1}{5}\sqrt{150}$.

d) Ta có: $\frac{1}{2}\sqrt{6}=\sqrt{{\left(\frac{1}{2}\right)}^2.6}=\sqrt{\frac{6}{4}}$

$=\sqrt{\frac{3}{2}}=\sqrt{3.\frac{1}{2}}=\sqrt{3}.\sqrt{\frac{1}{2}}$

Vì $\sqrt{3}.\sqrt{\frac{1}{2}}<6\sqrt{\frac{1}{2}}$ nên $\frac{1}{2}.\sqrt{6}<6\sqrt{\frac{1}{2}}$

Vậy $\frac{1}{2}\sqrt{6}<6\sqrt{\frac{1}{2}}$.

Bài 6: 

a: \(15=\sqrt{225}>\sqrt{200}\)

b: \(27=9\sqrt{9}>9\sqrt{5}\)

c: \(-24=-\sqrt{576}< -\sqrt{540}=-6\sqrt{15}\)

Bài 1. Tính căn bậc hai số học của các số sau:

1) 36=\(\sqrt{36}=4\)

2) 81\(\sqrt{81}=9\)

3) 121=\(\sqrt{121}=11\)

4) 144=\(\sqrt{144}=12\)

5) 0,16=\(\sqrt{0,16}=0,4\)

7) 29=\(\sqrt{29}~5,39\)

8) 0=\(\sqrt{0}=0\)

Bài 2: 

1: \(\sqrt{6}< \sqrt{41}\)

2: \(\sqrt{19}>\sqrt{4}\)

3: \(\sqrt{21}>\sqrt{5}\)

4: \(\sqrt{7}< \sqrt{51}\)

\(\sqrt{27}+\sqrt{6}+1< \sqrt{48}\)

Bạn chỉ mình cách làm đc k

ta có: \(\sqrt{27}+\sqrt{6}+1=3\sqrt{3}+\sqrt{6}+1\)(1))

          \(\sqrt{48}=4\sqrt{3}=3\sqrt{3}+\sqrt{3}\)(2)

ta lại có: \(\sqrt{6}>\sqrt{3}\Rightarrow\sqrt{6}+1>\sqrt{3}\) (3)

từ (1)(2)và(3)\(\Rightarrow3\sqrt{3}+\sqrt{6}+1>3\sqrt{3}+\sqrt{3}\)

                   \(\Leftrightarrow\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)