\( \begin{array}{ll}\text { 10 } \sqrt{25 a^{6} p^{10}} & \text { 2. } \sqrt[3]{8 x^{12} y^{15}} \\ \text { 3. } \sqrt[5]{32 x^{5} y^{10}} & \text { 4. } \sqrt{2 x^{3}} \times \sqrt{8 x^{5}} \\ \)\begin{tabular}{ll}\text { (5. }\( \sqrt{2} \times \sqrt{8} \) & \text { 6. }\( \sqrt[6]{64 a^{5} b^{12} c^{18}} \) \\ \text { (7) }\( \sqrt{9 x^{10}+16 x^{10}} \) & \text { 8. }\( \sqrt{\frac{27 x^{3} y^{5}}{12 x^{3} y^{3}}} \) \\ \text { 9. }\( \sqrt{3 m^{3}} \times \sqrt{27 m} \) & \text { 10. }\( \sqrt{2 m^{3}} \times \sqrt{50 m x^{2}} \) \\ \text { 11. }\( \frac{\sqrt{9 x^{10}}+\sqrt{4 x^{10}}}{5 x^{3}} \) & \text { 12. }\( \sqrt[3]{3 x} \times \sqrt[3]{72 x^{5}} \) \\ \hline\end{tabular}\end{array}

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: \[ \sqrt[3]{216x^6}=6x^2. \]