3:
a: \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)
b: \(\dfrac{x}{y}\cdot\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)
2:
a: 2căn 7=căn 28
3căn 2=căn 18
mà 28>18
nên 2*căn 7>3*căn 2
b: 5=2+3
mà 3>căn 2
nên 2+3>2+căn 2
=>5>2+căn 2
1) a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
\(=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}\)
\(=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}\)
\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49}\)
\(=3\sqrt{a}-4\sqrt{a}+7=7-\sqrt{a}\)
2. a) \(2\sqrt{7}=\sqrt{4.7}=\sqrt{28}\)
\(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)
Mà \(\sqrt{28}>\sqrt{18}\Rightarrow2\sqrt{7}>3\sqrt{2}\)
b) \(5=2+3=2+\sqrt{9}\)
Vì \(\sqrt{9}>\sqrt{2}\Rightarrow2+\sqrt{9}>2+\sqrt{2}\Rightarrow5>2+\sqrt{2}\)
3. a) \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)
b) \(\dfrac{x}{y}.\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}.\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)