Question

Can someone please help me?

Answer 1

For each square root, you can simplify the expression under the root as

$3\sqrt{98xy}=3\sqrt{2\cdot7^2xy}=3\cdot7\sqrt{2xy}=21\sqrt{2xy}$

$\sqrt{108y}=\sqrt{2^2\cdot3^3xy}=2\cdot3\sqrt{3xy}=6\sqrt{3xy}$

$2\sqrt{75xy}=2\sqrt{3\cdot5^2xy}=2\cdot5\sqrt{3xy}=10\sqrt{3xy}$

So we have

$3\sqrt{98xy}-\sqrt{108xy}+2\sqrt{75xy}=21\sqrt{2xy}-6\sqrt{3xy}+10\sqrt{3xy}$

$=21\sqrt{2xy}+4\sqrt{3xy}$

This means [1] = 4 and [2] = $3xy$.