Product, division and sum of square roots

Calculate the square roots of the following perfect squares:

$\sqrt{1521}$. Notice that $1521=9\cdot169=(3\cdot3)\cdot(13\cdot13)$, therefore we are sure that it is a perfect square.
$\sqrt{\dfrac{2916}{484}}$. Notice that $2916=36\cdot81$ and $494=121\cdot4$.
$\sqrt{64}+3\sqrt{81}$
$3\sqrt{19}-2\sqrt{49}+\dfrac{\sqrt{475}}{5}$ where $475=19\cdot25$

As is indicated, $1521=9\cdot169$ so we can write the root as $\sqrt{1521}=\sqrt{9\cdot169}$.

Now, using the first property of the square root, we have: $$\sqrt{1521}=\sqrt{9\cdot169}=\sqrt{9}\cdot\sqrt{169}=3\cdot13=39$$

Writing inside the root the information given in the statement, we have $\sqrt{\dfrac{2916}{484}}=\sqrt{\dfrac{36\cdot81}{121\cdot4}}$ and if we use the second property of the square root $$\sqrt{\dfrac{36\cdot81}{121\cdot4}}=\dfrac{\sqrt{36\cdot81}}{\sqrt{121\cdot4}}$$ Now we proceed as in the previous paragraph, using the first property of roots and: $$\dfrac{\sqrt{36\cdot81}}{\sqrt{121\cdot4}}=\dfrac{\sqrt{36}\sqrt{81}}{\sqrt{121}\sqrt{4}}=\dfrac{6\cdot9}{11\cdot4}=\dfrac{54}{44}$$
With the table of most common perfect squares we have $$\sqrt{64}+3\sqrt{81}=8+3\cdot9=8+27=35$$
We realise that we cannot calculate the root of $19$ because it is not a perfect square, therefore we just solve the other roots and then gather together the terms with a root on the one hand, and those that do not on the other $$3\sqrt{19}-2\sqrt{49}+\dfrac{\sqrt{475}}{5}=3\sqrt{19}-2\cdot7+\dfrac{\sqrt{25\cdot19}}{5}=$$ $$=3\sqrt{19}-2\cdot7+\dfrac{\sqrt{25}\cdot\sqrt{19}}{5}=3\sqrt{19}-14+\dfrac{5\cdot\sqrt{19}}{5}=$$ $$=(3+\dfrac{5}{5})\sqrt{19}-14=4\sqrt{19}-14$$