Asked by idk
simplify: index of 4 √36m^8n^4 where m≥0 , n≥0
Answers
Answered by
GPT-4o mini
To simplify the expression \( \sqrt[4]{36m^8n^4} \), we can break it down as follows:
1. **Identify Components Inside the Root:**
- \( 36 \) can be expressed as \( 6^2 \).
- \( m^8 \) is already a perfect power.
- \( n^4 \) is also a perfect power.
2. **Rewriting under the Fourth Root:**
\[
\sqrt[4]{36m^8n^4} = \sqrt[4]{6^2 \cdot m^8 \cdot n^4}
\]
3. **Applying the Fourth Root:**
- \( \sqrt[4]{6^2} \) simplifies to \( 6^{2/4} = 6^{1/2} = \sqrt{6} \).
- \( \sqrt[4]{m^8} = m^{8/4} = m^2 \).
- \( \sqrt[4]{n^4} = n^{4/4} = n \).
4. **Combining the Results:**
Putting everything together, we have:
\[
\sqrt[4]{36m^8n^4} = \sqrt{6} \cdot m^2 \cdot n
\]
Thus, the simplified expression is:
\[
m^2n\sqrt{6}
\]
This is the final simplified form of the given expression for \( m \geq 0, n \geq 0 \).
1. **Identify Components Inside the Root:**
- \( 36 \) can be expressed as \( 6^2 \).
- \( m^8 \) is already a perfect power.
- \( n^4 \) is also a perfect power.
2. **Rewriting under the Fourth Root:**
\[
\sqrt[4]{36m^8n^4} = \sqrt[4]{6^2 \cdot m^8 \cdot n^4}
\]
3. **Applying the Fourth Root:**
- \( \sqrt[4]{6^2} \) simplifies to \( 6^{2/4} = 6^{1/2} = \sqrt{6} \).
- \( \sqrt[4]{m^8} = m^{8/4} = m^2 \).
- \( \sqrt[4]{n^4} = n^{4/4} = n \).
4. **Combining the Results:**
Putting everything together, we have:
\[
\sqrt[4]{36m^8n^4} = \sqrt{6} \cdot m^2 \cdot n
\]
Thus, the simplified expression is:
\[
m^2n\sqrt{6}
\]
This is the final simplified form of the given expression for \( m \geq 0, n \geq 0 \).
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