Question

1) (4-c)(4+c)= 2) (7b+1)(1-7b)=

Answer 1

Ответ:

1) (4-с)(4+с)= 4^2-с^2=16-с^2

2) (7b+1)(1-7b)= (1+7b)(1-7b)= 1-49b^2

Answer 2

Let's solve each equation one by one:

1. (4 - c)(4 + c)

To simplify this expression, we can use the difference of squares identity, which states that (a - b)(a + b) = a^2 - b^2.

In this case, a = 4 and b = c. Therefore, we have:

(4 - c)(4 + c) = 4^2 - c^2 = 16 - c^2

So the simplified expression is 16 - c^2.

2. (7b + 1)(1 - 7b)

Similarly, we can use the difference of squares identity to simplify this expression.

In this case, a = 1 and b = 7b. Therefore, we have:

(7b + 1)(1 - 7b) = (1)^2 - (7b)^2 = 1 - 49b^2

So the simplified expression is 1 - 49b^2.