Question

[Kim Khánh Linh]

Bài 45 (trang 27 SGK Toán 9 Tập 1)

So sánh

a) $3 \sqrt{3}$ và $\sqrt{12}$ ;                            b) $7$ và $3 \sqrt{5}$ ;

c) $\dfrac{1}{3} \sqrt{51}$ và $\dfrac{1}{5} \sqrt{150}$ ;                   d) $\dfrac{1}{2} \sqrt{6}$ và $6 \sqrt{\dfrac{1}{2}}$.

Answer 1

[Ngọc Mai_NBK]

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\(\sqrt{\frac{1}{2}}\)=\(\sqrt{\frac{36}{2}}\)=\(\sqrt{18}\)

Answer 2

[Phạm Bá Huy]

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