Question

1. \\(sqrt{98n^{12}}\\)
2. \\(sqrt{28k^9}\\)
3. \\(sqrt{18x^4y}\\)
4. \\(sqrt{24mn^2}\\)
5. \\(sqrt{50x^3y^4}\\)
6. \\(sqrt{320mn^4}\\)
7. \\(sqrt{192a^3b}\\)
8. \\(sqrt{80x^2y^4}\\)
9. \\(sqrt{112x^3y^4}\\)
10. \\(sqrt{256x^4y^4}\\)
11. \\(sqrt{216x^3y^3}\\)
12. \\(sqrt{192u^4v^4}\\)

Explanation:

Let's solve problem 19: $\sqrt{98n^{12}}$

Step 1: Factor the radicand

We can factor 98 into $49\times2$ and $n^{12}$ is a perfect square. So, $\sqrt{98n^{12}}=\sqrt{49\times2\times n^{12}}$

Step 2: Use the property of square roots

We know that $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ (for $a\geq0, b\geq0$). So we can split the square root:
$\sqrt{49\times2\times n^{12}}=\sqrt{49}\times\sqrt{2}\times\sqrt{n^{12}}$

Step 3: Simplify each square root

We know that $\sqrt{49} = 7$, and $\sqrt{n^{12}}=n^{6}$ (since $(n^{6})^{2}=n^{12}$). So putting it together:
$7\times\sqrt{2}\times n^{6}=7n^{6}\sqrt{2}$

Answer:

$7n^{6}\sqrt{2}$