Câu 45:

So sánh:

a) \(3 \sqrt{3}\) và \(\sqrt{12}\)

b) 7 và \(3 \sqrt{5}\)

c) \(\frac{1}{3} \sqrt{51}\) và \(\frac{1}{5} \sqrt{150}\)

d) \(\frac{1}{2} \sqrt{6}\) và \(6 \sqrt{\frac{1}{2}}\)

Lời giải:

a) \(3 \sqrt{3}=\sqrt{3^{2} .3}=\sqrt{9.3}=\sqrt{27}>\sqrt{12}\)

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 \(\sqrt{49}>\sqrt{45} \text { nên } 7>3 \sqrt{5}\)

c) Ta có: \(\frac{1}{3} \sqrt{51}=\sqrt{\left(\frac{1}{3}\right)^{2} .51}=\sqrt{\frac{51}{9}}\)

\(\frac{1}{5} \sqrt{150}=\sqrt{\left(\frac{1}{5}\right)^{2} \cdot 150}=\sqrt{\frac{150}{25}}=\sqrt{6}\)

Do đó: \(\frac{1}{5} \sqrt{150}=\sqrt{6}=\sqrt{\frac{6.9}{9}}=\sqrt{\frac{54}{9}}>\sqrt{\frac{51}{9}}\)

Vậy \(\frac{1}{3} \sqrt{51}<\frac{1}{5} \sqrt{150}\)

d) \(\frac{1}{2} \sqrt{6}=\sqrt{\left(\frac{1}{2}\right)^{2} \cdot 6}=\sqrt{\frac{6}{4}}=\sqrt{\frac{3}{2}}=\sqrt{3 \cdot \frac{1}{2}} =\sqrt{3} \cdot \sqrt{\frac{1}{2}}<6 \sqrt{\frac{1}{2}}\)

Vậy \(\frac{1}{2} \cdot \sqrt{6}<6 \sqrt{\frac{1}{2}}\)