174 Запишите A + B, A - B и B - A как многочлены в стандартном виде, если: a) A = 4 - 2xy + 5x² - 3y², B = 4x² – 3xy + 2y² - 2; б) A = 3a² - 5ab – (b² – 2), B = 5a² + 7ab + 1 – 3b².

a) Дано:

$$A = 4 - 2xy + 5x^2 - 3y^2$$

$$B = 4x^2 - 3xy + 2y^2 - 2$$

1) Найдем A + B:

$$A + B = (4 - 2xy + 5x^2 - 3y^2) + (4x^2 - 3xy + 2y^2 - 2) =$$

$$= 4 - 2xy + 5x^2 - 3y^2 + 4x^2 - 3xy + 2y^2 - 2 = $$

$$= (5x^2 + 4x^2) + (-2xy - 3xy) + (-3y^2 + 2y^2) + (4 - 2) = $$

$$= 9x^2 - 5xy - y^2 + 2$$

2) Найдем A - B:

$$A - B = (4 - 2xy + 5x^2 - 3y^2) - (4x^2 - 3xy + 2y^2 - 2) =$$

$$= 4 - 2xy + 5x^2 - 3y^2 - 4x^2 + 3xy - 2y^2 + 2 =$$

$$= (5x^2 - 4x^2) + (-2xy + 3xy) + (-3y^2 - 2y^2) + (4 + 2) =$$

$$= x^2 + xy - 5y^2 + 6$$

3) Найдем B - A:

$$B - A = (4x^2 - 3xy + 2y^2 - 2) - (4 - 2xy + 5x^2 - 3y^2) =$$

$$= 4x^2 - 3xy + 2y^2 - 2 - 4 + 2xy - 5x^2 + 3y^2 =$$

$$= (4x^2 - 5x^2) + (-3xy + 2xy) + (2y^2 + 3y^2) + (-2 - 4) =$$

$$= -x^2 - xy + 5y^2 - 6$$

Ответ: A + B = $$9x^2 - 5xy - y^2 + 2$$, A - B = $$x^2 + xy - 5y^2 + 6$$, B - A = $$-x^2 - xy + 5y^2 - 6$$

б) Дано:

$$A = 3a^2 - 5ab - (b^2 - 2) = 3a^2 - 5ab - b^2 + 2$$

$$B = 5a^2 + 7ab + 1 - 3b^2$$

1) Найдем A + B:

$$A + B = (3a^2 - 5ab - b^2 + 2) + (5a^2 + 7ab + 1 - 3b^2) =$$

$$= 3a^2 - 5ab - b^2 + 2 + 5a^2 + 7ab + 1 - 3b^2 =$$

$$= (3a^2 + 5a^2) + (-5ab + 7ab) + (-b^2 - 3b^2) + (2 + 1) =$$

$$= 8a^2 + 2ab - 4b^2 + 3$$

2) Найдем A - B:

$$A - B = (3a^2 - 5ab - b^2 + 2) - (5a^2 + 7ab + 1 - 3b^2) =$$

$$= 3a^2 - 5ab - b^2 + 2 - 5a^2 - 7ab - 1 + 3b^2 =$$

$$= (3a^2 - 5a^2) + (-5ab - 7ab) + (-b^2 + 3b^2) + (2 - 1) =$$

$$= -2a^2 - 12ab + 2b^2 + 1$$

3) Найдем B - A:

$$B - A = (5a^2 + 7ab + 1 - 3b^2) - (3a^2 - 5ab - b^2 + 2) =$$

$$= 5a^2 + 7ab + 1 - 3b^2 - 3a^2 + 5ab + b^2 - 2 =$$

$$= (5a^2 - 3a^2) + (7ab + 5ab) + (-3b^2 + b^2) + (1 - 2) =$$

$$= 2a^2 + 12ab - 2b^2 - 1$$

Ответ: A + B = $$8a^2 + 2ab - 4b^2 + 3$$, A - B = $$-2a^2 - 12ab + 2b^2 + 1$$, B - A = $$2a^2 + 12ab - 2b^2 - 1$$