\(\rm (\sqrt{98}-\sqrt{72}+\sqrt{50})\div\sqrt{18}=?\)

Given:

\((\sqrt{98}-\sqrt{72}+\sqrt{50})\div\sqrt{18}\) = ?

Formula used:

To simplify, use the property of square roots: \(\sqrt{a\times b} = \sqrt{a}\times\sqrt{b}\)

Calculation:

\(\sqrt{98}\) = \(\sqrt{2 \times 49} = \sqrt{2} \times 7 = 7\sqrt{2}\)2 $\sqrt{2} \times 7 = 7\sqrt{2}$

\(\sqrt{72}\) = \(\sqrt{2 \times 36} = \sqrt{2} \times 6 = 6\sqrt{2}\)2 $\sqrt{2} \times 6 = 6\sqrt{2}$

\(\sqrt{50}\) = \(\sqrt{2 \times 25} = \sqrt{2} \times 5 = 5\sqrt{2}\)2 $\sqrt{2} \times 5 = 5\sqrt{2}$

\(\sqrt{18}\) = \(\sqrt{2 \times 9} = \sqrt{2} \times 3 = 3\sqrt{2}\)2 $\sqrt{2} \times 3 = 3\sqrt{2}$

⇒ \((7\sqrt{2} - 6\sqrt{2} + 5\sqrt{2})\div 3\sqrt{2}\)

⇒ \((6\sqrt{2})\div 3\sqrt{2}\)

⇒ 2

∴ The correct answer is option (1).