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Answer:

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Step-by-step explanation:

1. To solve 3a² - 66a + 363 for a perfect square trinomial, we need to take half of the coefficient of the middle term, square it, and add it to both sides of the equation. The coefficient of the squared term is 3, and half of the coefficient of the middle term is -33/2. Squaring -33/2 gives us 1089/4.

3a² - 66a + 363 + 1089/4 - 1089/4 = (a - 11)²

Simplifying the equation, we get:

3a² - 66a + 1458/4 = (a - 11)²

Therefore, the perfect square trinomial for 3a² - 66a + 363 is (a - 11)²

2. To solve (a+1)c² - 8c(a+1) + 16(a+1) for a perfect square trinomial, we need to complete the square for the first two terms. Taking half of the coefficient of the middle term, we get:

(a+1)c² - 8c(a+1) + 16(a+1) = (a+1)(c-4)²

Therefore, the perfect square trinomial for (a+1)c² - 8c(a+1) + 16(a+1) is (a+1)(c-4)².

3. To solve 100(a-4) + 20b(a-4) + b'(a²-4) for a perfect square trinomial, we can factor out (a-4) from the first two terms and complete the square for the third term. Taking half of the coefficient of the squared term, we get:

100(a-4) + 20b(a-4) + b'(a²-4) = (a-4)(100+20b+b'a²)

100+20b+b'a² = (b'a)²/4 + 10b'a + 100

Substituting into the original equation, we get:

100(a-4) + 20b(a-4) + b'(a²-4) = (a-4)((b'a)²/4 + 10b'a + 100)

Therefore, the perfect square trinomial for 100(a-4) + 20b(a-4) + b'(a²-4) is (b'a/2 + 10)(a-4)².

4. To solve 4cd² - 20cd³ + 25d" for a perfect square trinomial, we can factor out d² and complete the square for the remaining terms. Taking half of the coefficient of the middle term, we get:

4cd² - 20cd³ + 25d" = d²(4c - 20cd + 25d)

4c - 20cd + 25d = 5(2 - 2cd + 5d) = 5(1 - cd + 5d)²

Substituting into the original equation, we get:

4cd² - 20cd³ + 25d" = d²(5(1 - cd + 5d)²)

Therefore, the perfect square trinomial for 4cd² - 20cd³ + 25d" is d²(5(1 - cd + 5d)²)

5. To solve 4(a-10ab+ 25b)-81c(al-10ub + 25b) for a perfect square trinomial, we can factor out the common factor of (a-5b)² from both terms. Simplifying, we get:

4(a-10ab+ 25b)-81c(al-10ub + 25b) = (a-5b)²(4-81c)

Therefore, the perfect square trinomial for 4(a-10ab+ 25b)-81c(al-10ub + 25b) is (a-5b)²(4-81c).

Answer:

To determine which of the given trinomials are perfect squares, we can try to factor them and check if they can be written in the form (x + y)².

1. 3a² - 66a + 363

To factor this trinomial, we can find two numbers that multiply to give 363 and add/subtract to give -66. However, 363 is not a perfect square and cannot be factored evenly. Therefore, this trinomial is not a perfect square.

2. (a + 1)c² - 8c(a + 1) + 16(a + 1)

This trinomial can be factored as (a + 1)². Therefore, it is a perfect square.

3. 100(a² - 4) + 20b(a² - 4) + b²(a² - 4)

We can factor out (a² - 4) from each term, giving us (a² - 4)(100 + 20b + b²). This trinomial is not a perfect square.

4. 4c²d² - 20cd³ + 25d⁴

This trinomial can be factored as (2cd - 5d²)². Therefore, it is a perfect square.

5. 4(a² + 10ab + 25b²) - 81c² (a² - 10ab + 25b²)

This trinomial cannot be factored into a perfect square since the second term has a negative sign. Therefore, this trinomial is not a perfect square.

In conclusion, the perfect square trinomials are:

- (a + 1)c² - 8c(a + 1) + 16(a + 1)

- 4c²d² - 20cd³ + 25d⁴