Упр.32.15 Часть 2 ГДЗ Мордкович Семенов 7 класс (Алгебра)

P_1 (a; b) = a^2 + 2ab — b^2,
p_2 (a; b) = 3a^2 — ab + b^2,
p_3 (a; b) = —a^2 + ab — 3b^2.
Найдите:
а) p(a; b) = p_1 (a; b) — p_2 (a; b) + p_3 (a; b);
б) p(a; b) = p_2 (a; b) — p_1 (a; b) — p_3 (a; b);
в) p(a; b) = p_3 (a; b) — p_2 (a; b) — p_1 (a; b);
г) p(a; b) = p_1 (a; b) + p_2 (a; b) — p_3 (a; b).

Даны:

$$p_1(a;b)=a^2+2ab-b^2,$$

$$p_2(a;b)=3a^2-ab+b^2,$$

$$p_3(a;b)=-a^2+ab-3b^2.$$

1. $$ \begin{aligned} p(a;b)&=p_1(a;b)-p_2(a;b)+p_3(a;b)\\ &=(a^2+2ab-b^2)-(3a^2-ab+b^2)+(-a^2+ab-3b^2)\\ &=a^2+2ab-b^2-3a^2+ab-b^2-a^2+ab-3b^2\\ &=-3a^2+4ab-5b^2. \end{aligned} $$

2. $$ \begin{aligned} p(a;b)&=p_2(a;b)-p_1(a;b)-p_3(a;b)\\ &=(3a^2-ab+b^2)-(a^2+2ab-b^2)-(-a^2+ab-3b^2)\\ &=3a^2-ab+b^2-a^2-2ab+b^2+a^2-ab+3b^2\\ &=3a^2-4ab+5b^2. \end{aligned} $$

3. $$ \begin{aligned} p(a;b)&=p_3(a;b)-p_2(a;b)-p_1(a;b)\\ &=(-a^2+ab-3b^2)-(3a^2-ab+b^2)-(a^2+2ab-b^2)\\ &=-a^2+ab-3b^2-3a^2+ab-b^2-a^2-2ab+b^2\\ &=-5a^2-3b^2. \end{aligned} $$

4. $$ \begin{aligned} p(a;b)&=p_1(a;b)+p_2(a;b)-p_3(a;b)\\ &=(a^2+2ab-b^2)+(3a^2-ab+b^2)-(-a^2+ab-3b^2)\\ &=a^2+2ab-b^2+3a^2-ab+b^2+a^2-ab+3b^2\\ &=5a^2+3b^2. \end{aligned} $$

Ответ

а) $$-3a^2+4ab-5b^2$$; б) $$3a^2-4ab+5b^2$$; в) $$-5a^2-3b^2$$; г) $$5a^2+3b^2$$.