Problems from Product, division and sum of square roots

Development:

1. As is indicated, $1521=9\cdot 169$ so we can write the root as $\sqrt{1521}=\sqrt{9\cdot 169}$.

Now, using the first property of the square root, we have: $$\sqrt{1521}=\sqrt{9\cdot 169}=\sqrt{9}\cdot \sqrt{169}=3\cdot 13=39$$

1. Writing inside the root the information given in the statement, we have $\sqrt{{\displaystyle \frac{2916}{484}}}=\sqrt{{\displaystyle \frac{36\cdot 81}{121\cdot 4}}}$ and if we use the second property of the square root $$\sqrt{{\displaystyle \frac{36\cdot 81}{121\cdot 4}}}={\displaystyle \frac{\sqrt{36\cdot 81}}{\sqrt{121\cdot 4}}}$$ Now we proceed as in the previous paragraph, using the first property of roots and: $${\displaystyle \frac{\sqrt{36\cdot 81}}{\sqrt{121\cdot 4}}}={\displaystyle \frac{\sqrt{36}\sqrt{81}}{\sqrt{121}\sqrt{4}}}={\displaystyle \frac{6\cdot 9}{11\cdot 4}}={\displaystyle \frac{54}{44}}$$

2. With the table of most common perfect squares we have $$\sqrt{64}+3\sqrt{81}=8+3\cdot 9=8+27=35$$

3. 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 $$3\sqrt{19}-2\sqrt{49}+{\displaystyle \frac{\sqrt{475}}{5}}=3\sqrt{19}-2\cdot 7+{\displaystyle \frac{\sqrt{25\cdot 19}}{5}}=$$ $$=3\sqrt{19}-2\cdot 7+{\displaystyle \frac{\sqrt{25}\cdot \sqrt{19}}{5}}=3\sqrt{19}-14+{\displaystyle \frac{5\cdot \sqrt{19}}{5}}=$$ $$=(3+{\displaystyle \frac{5}{5}})\sqrt{19}-14=4\sqrt{19}-14$$

Solution:

1. $39$
2. ${\displaystyle \frac{54}{44}}$
3. $35$
4. $4\sqrt{19}-14$

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