Question

698. a) a2 + 2am + m²;B) a2 + 4ab + 4b2;6) x2 + 4x + 4;r) b2 - 6b+9.​

Answer 1

a) Это формула квадратов (a+m)^2

b) (a+2b)^2

c) (x+2)^2

d) (b-3)^2

Answer 2

Ответ:

а) $>a + m<img src=$

б) $>a + 2b<img src=$

в) $>x + 2<img src=$

г) $>b - 3<img src=$

Объяснение:

а) a2 + 2am + m2 = $>a + m<img src=$

б) a2 + 4ab + 4b2 = $>a + 2b<img src=$

в) x2 + 4x + 4 = $>x + 2<img src=$

г) b2 - 6b + 9 = $>b - 3<img src=$

Answer 3

It looks like you've provided four expressions, and I assume you want to factor each of them. Let's factor each expression:

a) a^2 + 2am + m^2 This expression can be factored as a perfect square trinomial: a^2 + 2am + m^2 = (a + m)^2

b) a^2 + 4ab + 4b^2 This expression is also a perfect square trinomial: a^2 + 4ab + 4b^2 = (a + 2b)^2

6. x^2 + 4x + 4 This is also a perfect square trinomial: x^2 + 4x + 4 = (x + 2)^2

r) b^2 - 6b + 9 This expression is a perfect square trinomial as well: b^2 - 6b + 9 = (b - 3)^2

So, each of the given expressions is a perfect square trinomial and can be factored as shown above.