Question

A) (c-2)*(c+3)-(c-1)²= б) 3*(a+c)²-6ac=

Answer 1

Ответ:

a)3c-7

b)3(a^2+c^2)

Объяснение:

Answer 2

It looks like you have provided two mathematical expressions, A) and B), and you might be looking for their simplified forms. Let's work on simplifying each expression:

A) (c - 2) * (c + 3) - (c - 1)²

First, let's expand the expressions within the parentheses: (c - 2) * (c + 3) = c² + 3c - 2c - 6 = c² + c - 6

(c - 1)² = c² - 2c + 1

Now, substitute these expanded expressions back into the original equation: (c² + c - 6) - (c² - 2c + 1)

Now, distribute the negative sign to both terms in the second set of parentheses: c² + c - 6 - c² + 2c - 1

Combine like terms: 3c - 7

So, the simplified form of A) is: 3c - 7

B) 3 * (a + c)² - 6ac

First, square the binomial (a + c)²: (a + c)² = a² + 2ac + c²

Now, substitute this back into the original equation: 3 * (a² + 2ac + c²) - 6ac

Distribute the 3 to all terms inside the parentheses: 3a² + 6ac + 3c² - 6ac

Notice that the 6ac terms cancel out: 3a² + 3c²

So, the simplified form of B) is: 3a² + 3c²

If you have any further questions or need more assistance, feel free to ask!