Question

Выполните указанные действия 1)(m+n)(m-n) 1)(m+n)(m-n) 2)(a-3)(3+a) 3)(a-3)(3+a) 4)(y+2)(2-y)

Answer 1

Формулы сокращённого умножения. Разность квадратов

Answer 2

Давайте выполним указанные действия:

1) (m+n)(m-n): To expand this expression, we can use the FOIL method, which stands for First, Outer, Inner, Last. We multiply the first terms, then the outer terms, then the inner terms, and finally the last terms. Applying this method, we get: (m+n)(m-n) = m*m - m*n + n*m - n*n Simplifying further, we have: m^2 - mn + mn - n^2 The middle terms, -mn and mn, cancel each other out, leaving us with: m^2 - n^2.

2) (a-3)(3+a): Again, we can use the FOIL method to expand this expression: (a-3)(3+a) = a*3 + a*a - 3*a - 3*3 Simplifying further, we have: 3a + a^2 - 3a - 9 The terms 3a and -3a cancel each other out, leaving us with: a^2 - 9.

3) (a-3)(3+a): Using the FOIL method, we can expand this expression: (a-3)(3+a) = a*3 + a*a - 3*a - 3*3 Simplifying further, we have: 3a + a^2 - 3a - 9 The terms 3a and -3a cancel each other out, leaving us with: a^2 - 9.

4) (y+2)(2-y): Once again, we can use the FOIL method to expand this expression: (y+2)(2-y) = y*2 + y*(-y) + 2*2 + 2*(-y) Simplifying further, we have: 2y - y^2 + 4 - 2y The terms 2y and -2y cancel each other out, leaving us with: -y^2 + 4.

So, the expanded forms of the given expressions are: 1) (m+n)(m-n) = m^2 - n^2 2) (a-3)(3+a) = a^2 - 9 3) (a-3)(3+a) = a^2 - 9 4) (y+2)(2-y) = -y^2 + 4.

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