Question

sqrt(243/867)+sqrt(1183/2023)

Answer 1

To simplify the expression \(\sqrt{\frac{243}{867}} + \sqrt{\frac{1183}{2023}}\), we can first simplify each square root separately.

1. **Simplifying \(\sqrt{\frac{243}{867}}\)**:

- Factor \(243\) and \(867\).
- \(243 = 3^5\)
- \(867 = 3 \times 289 = 3 \times 17^2\)

Thus:

\[
\frac{243}{867} = \frac{3^5}{3 \times 17^2} = \frac{3^4}{17^2} = \frac{81}{289}
\]

Now, we can take the square root:

\[
\sqrt{\frac{243}{867}} = \sqrt{\frac{81}{289}} = \frac{\sqrt{81}}{\sqrt{289}} = \frac{9}{17}
\]

2. **Simplifying \(\sqrt{\frac{1183}{2023}}\)**:

- First, check if \(1183\) and \(2023\) can be factored easily.
- \(1183\) is a prime number.
- To factor \(2023\):

- Check divisibility by \(3\): \(2 + 0 + 2 + 3 = 7\) (not divisible).
- Check divisibility by \(7\): \(2023 \div 7 \approx 289\), so \(2023 = 7 \times 289\).
- \(289 = 17^2\), thus \(2023 = 7 \times 17^2\).

Now, we can write:

\[
\frac{1183}{2023} = \frac{1183}{7 \times 17^2}
\]

Taking the square root:

\[
\sqrt{\frac{1183}{2023}} = \sqrt{\frac{1183}{7 \times 17^2}} = \frac{\sqrt{1183}}{17\sqrt{7}}
\]

Since there are no perfect squares in \(1183\) and it remains in the radical, we cannot simplify further.

Putting it all together:

\[
\sqrt{\frac{243}{867}} + \sqrt{\frac{1183}{2023}} = \frac{9}{17} + \frac{\sqrt{1183}}{17\sqrt{7}}
\]

Combining the terms gives:

\[
\frac{9 + \frac{\sqrt{1183}}{\sqrt{7}}}{17}
\]

Thus, the simplified expression is:

\[
\frac{9 + \sqrt{\frac{1183}{7}}}{17}
\]

This is the simplified result for the expression \( \sqrt{\frac{243}{867}} + \sqrt{\frac{1183}{2023}} \).