Question

(a-c)(a+c)-c(3a-c) (a-3)(a-7)-2a(3a-5) решите........

Answer 1

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Answer 2

To solve the expressions, let's start by simplifying each of them step by step:

1. (a-c)(a+c) - c(3a-c) Expanding the first part: (a - c)(a + c) = a(a) + a(c) - c(a) - c(c) = a^2 + ac - ac - c^2 = a^2 - c^2

Now, let's simplify the second part: c(3a - c) = 3ac - c^2

Now, combine the two simplified parts: (a^2 - c^2) - (3ac - c^2)

Notice that (a^2 - c^2) and (- c^2) cancel out: (a^2 - c^2) - (3ac - c^2) = a^2 - 3ac

2. (a - 3)(a - 7) - 2a(3a - 5) Expanding the first part: (a - 3)(a - 7) = a(a) - 7a - 3(a) + 21 = a^2 - 10a + 21

Now, simplify the second part: 2a(3a - 5) = 6a^2 - 10a

Now, combine the two simplified parts: (a^2 - 10a + 21) - (6a^2 - 10a)

Distribute the negative sign in the second part: (a^2 - 10a + 21) - 6a^2 + 10a

Combine like terms: a^2 - 6a^2 - 10a + 10a + 21

Simplify further: -5a^2 + 21

So, the final results are:

1. (a - c)(a + c) - c(3a - c) = a^2 - 3ac
2. (a - 3)(a - 7) - 2a(3a - 5) = -5a^2 + 21