Answer
1. \( 50a^3p^5 \)
2. \( 2x^4y^5 \)
3. \( 2xy^2 \)
4. \( 4x^4 \)
5. \( 4 \)
6. \( 2b^2c^3\sqrt[6]{a^5} \)
7. \( 5x^5 \)
8. \( \frac{3y}{2} \)
9. \( 9m^2 \)
10. \( 10m^2x \)
11. \( x^2 \)
12. \( 6x^2 \)
Solution
1. Simplify
\[
10 \sqrt{25a^6 p^{10}}.
\]
Inside the square root, factor:
\[
25a^6p^{10} = 25\,(a^6)(p^{10}).
\]
Taking the square root gives:
\[
\sqrt{25}=5,\quad \sqrt{a^6}=a^3,\quad \sqrt{p^{10}}=p^5.
\]
Thus,
\[
10 \sqrt{25a^6p^{10}} = 10\,(5a^3p^5)=50a^3p^5.
\]
2. Simplify
\[
\sqrt[3]{8x^{12}y^{15}}.
\]
Factor each term:
\[
8=2^3,\quad x^{12}=(x^4)^3,\quad y^{15}=(y^5)^3.
\]
Taking the cube root yields:
\[
\sqrt[3]{2^3}=2,\quad \sqrt[3]{(x^4)^3}=x^4,\quad \sqrt[3]{(y^5)^3}=y^5.
\]
Therefore,
\[
\sqrt[3]{8x^{12}y^{15}}=2x^4y^5.
\]
3. Simplify
\[
\sqrt[5]{32x^5y^{10}}.
\]
Write each component as a fifth power:
\[
32=2^5,\quad x^5=x^5,\quad y^{10}=(y^2)^5.
\]
Then,
\[
\sqrt[5]{2^5}=2,\quad \sqrt[5]{x^5}=x,\quad \sqrt[5]{(y^2)^5}=y^2.
\]
Thus,
\[
\sqrt[5]{32x^5y^{10}}=2xy^2.
\]
4. Simplify
\[
\sqrt{2x^3}\times\sqrt{8x^5}.
\]
Combine under a single square root:
\[
\sqrt{2x^3}\times\sqrt{8x^5}=\sqrt{(2x^3)(8x^5)}=\sqrt{16x^8}.
\]
Since
\[
\sqrt{16}=4\quad \text{and}\quad \sqrt{x^8}=x^4,
\]
we obtain
\[
\sqrt{16x^8}=4x^4.
\]
5. Simplify
\[
\sqrt{2}\times\sqrt{8}.
\]
Combine the radicals:
\[
\sqrt{2}\times\sqrt{8}=\sqrt{16}=4.
\]
6. Simplify
\[
\sqrt[6]{64a^5b^{12}c^{18}}.
\]
Factor each term:
\[
64=2^6,\quad b^{12}=(b^2)^6,\quad c^{18}=(c^3)^6.
\]
The term \(a^5\) does not form a perfect sixth power, so it remains under the radical. Thus,
\[
\sqrt[6]{64a^5b^{12}c^{18}}=\sqrt[6]{2^6}\,\sqrt[6]{a^5}\,\sqrt[6]{(b^2)^6}\,\sqrt[6]{(c^3)^6}=2\,b^2\,c^3\,\sqrt[6]{a^5}.
\]
7. Simplify
\[
\sqrt{9x^{10}+16x^{10}}.
\]
First, combine like terms:
\[
9x^{10}+16x^{10}=(9+16)x^{10}=25x^{10}.
\]
Then,
\[
\sqrt{25x^{10}}=\sqrt{25}\,\sqrt{x^{10}}=5x^5.
\]
8. Simplify
\[
\sqrt{\frac{27x^3y^5}{12x^3y^3}}.
\]
Cancel the common factors \(x^3\) and \(y^3\):
\[
\frac{27x^3y^5}{12x^3y^3}=\frac{27y^2}{12}.
\]
Simplify the fraction:
\[
\frac{27}{12}=\frac{9}{4}.
\]
Thus,
\[
\sqrt{\frac{9y^2}{4}}=\frac{\sqrt{9y^2}}{\sqrt{4}}=\frac{3y}{2}.
\]
9. Simplify
\[
\sqrt{3m^3}\times\sqrt{27m}.
\]
Multiply under the radical:
\[
\sqrt{3m^3}\times\sqrt{27m}=\sqrt{(3m^3)(27m)}=\sqrt{81m^4}.
\]
Since
\[
\sqrt{81}=9,\quad \sqrt{m^4}=m^2,
\]
we have
\[
\sqrt{81m^4}=9m^2.
\]
10. Simplify
\[
\sqrt{2m^3}\times\sqrt{50mx^2}.
\]
Multiply under the square root:
\[
(2m^3)(50mx^2)=100m^4x^2.
\]
So,
\[
\sqrt{100m^4x^2}=\sqrt{100}\,\sqrt{m^4}\,\sqrt{x^2}=10m^2x.
\]
11. Simplify
\[
\frac{\sqrt{9x^{10}}+\sqrt{4x^{10}}}{5x^3}.
\]
Evaluate each square root:
\[
\sqrt{9x^{10}}=3x^5,\quad \sqrt{4x^{10}}=2x^5.
\]
Then, the numerator becomes:
\[
3x^5+2x^5=5x^5.
\]
Thus,
\[
\frac{5x^5}{5x^3}=x^{5-3}=x^2.
\]
12. Simplify
\[
\sqrt[3]{3x}\times\sqrt[3]{72x^5}.
\]
Combine under a single cube root:
\[
\sqrt[3]{3x}\times\sqrt[3]{72x^5}=\sqrt[3]{(3x)(72x^5)}=\sqrt[3]{216x^6}.
\]
Since
\[
216=6^3\quad \text{and}\quad x^6=(x^2)^3,
\]
we have:
\[
\sqrt[3]{216x^6}=6x^2.
\]
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