Let's simplify each of the given expressions step by step to ensure clarity:

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a) Simplify √150

We factorize 150 to simplify the square root:

1. Factorize 150 into prime factors:
   $150 = 2 \cdot 3 \cdot 5 \cdot 5$

2. Group perfect squares:
   $\sqrt{150} = \sqrt{5^2 \cdot 3 \cdot 2}$

3. Simplify the perfect square ($5^2$) to $5$:
   $\sqrt{150} = 5\sqrt{6}$

Final Answer:
$\sqrt{150} = 5\sqrt{6}$

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b) Simplify $O2\sqrt{10}PO5\sqrt{2}P$

Let’s interpret this as the product of two terms: $O2\sqrt{10}$ and $O5\sqrt{2}$, where $O$ and $P$ are placeholders/constants.

1. Simplify the square roots and multiply:
   $O2\sqrt{10} \cdot O5\sqrt{2} = (O2 \cdot O5)(\sqrt{10} \cdot \sqrt{2})$

2. Combine constants and simplify roots:
   $O2 \cdot O5 = O^2 \cdot 10$
   $\sqrt{10} \cdot \sqrt{2} = \sqrt{20} = 2\sqrt{5}$

3. Multiply the results:
   $O^2 \cdot 10 \cdot 2\sqrt{5} = 20O^2\sqrt{5}$

Final Answer:
$O2\sqrt{10}PO5\sqrt{2}P = 20O^2\sqrt{5}$

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c) Simplify $3\sqrt{48} - 5\sqrt{27} + 2\sqrt{3} - \sqrt{32}$

1. Simplify each term:

   • $\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$, so $3\sqrt{48} = 3 \cdot 4\sqrt{3} = 12\sqrt{3}$
   • $\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$, so $-5\sqrt{27} = -5 \cdot 3\sqrt{3} = -15\sqrt{3}$
   • $2\sqrt{3}$ remains as $2\sqrt{3}$
   • $\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}$, so $-\sqrt{32} = -4\sqrt{2}$

2. Combine like terms:

   • For $\sqrt{3}$ terms:
     $12\sqrt{3} - 15\sqrt{3} + 2\sqrt{3} = -1\sqrt{3}$
   • The $\sqrt{2}$ term is $-4\sqrt{2}$.

Final Answer:
[ 3\sqrt{48} - 5\sqrt{27