Упр.318 ГДЗ Никольский Потапов 7 класс (Алгебра)

Р±) (2a2bc — 3b2c — 7bc2) (a2c — b3c2 + 3bc3 — 8c2);
РІ) (m2 — mn2 — mn — n2) (m — mn — n2 + n);
r) (0,1p3 — 2p2q — 0,5pq2 + l,2p3)(8p2 — 0,2pq + 5q2).

а)

$$ (x^2+y^2+x+y)(x+y+xy)= $$
$$ =x^3+x^2y+x^3y+xy^2+y^3+xy^3+x^2+xy+x^2y+xy^2+y^2+xy^2 $$
$$ =x^3+2x^2y+x^3y+2xy^2+y^3+xy^3+x^2+2xy+y^2. $$

б)

$$ (2a^2bc-3b^2c-7bc^2)(a^2c-b^3c^2+3bc^3-8c^2)= $$
$$ =2a^4bc^2-2a^2b^4c^3+6a^2b^2c^4-16a^2bc^3 -3a^2b^2c^2+3b^5c^3-9b^3c^4+24b^2c^3 $$
$$ -7a^2bc^3+7b^4c^4-21b^2c^5+56bc^4. $$

в)

$$ (m^2-mn^2-mn-n^2)(m-mn-n^2+n)= $$
$$ =m^3-m^3n-m^2n^2+mn^2-m^2n^2+m^2n^3+mn^4-mn^3 $$
$$ -mn^2+mn^3+n^4-n^3 $$
$$ =m^3-m^3n-m^2n^2+m^2n^3+mn^4+mn^3-2mn^2+n^4-n^3. $$

г)

$$ (0{,}1p^3-2p^2q-0{,}5pq^2+1{,}2p^3)(8p^2-0{,}2pq+5q^2)= $$
$$ =(1{,}3p^3-2p^2q-0{,}5pq^2)(8p^2-0{,}2pq+5q^2)= $$
$$ =10{,}4p^5-0{,}26p^4q+6{,}5p^3q^2-16p^4q+0{,}4p^3q^2-10p^2q^3 $$
$$ -4p^3q^2+0{,}1p^2q^3-2{,}5pq^4 $$
$$ =10{,}4p^5-16{,}26p^4q+2{,}9p^3q^2-9{,}9p^2q^3-2{,}5pq^4. $$

Ответ

а) $$x^3+2x^2y+x^3y+2xy^2+y^3+xy^3+x^2+2xy+y^2$$;
б) $$2a^4bc^2-2a^2b^4c^3+6a^2b^2c^4-16a^2bc^3-3a^2b^2c^2+3b^5c^3-9b^3c^4+24b^2c^3-7a^2bc^3+7b^4c^4-21b^2c^5+56bc^4$$;
в) $$m^3-m^3n-m^2n^2+m^2n^3+mn^4+mn^3-2mn^2+n^4-n^3$$;
г) $$10{,}4p^5-16{,}26p^4q+2{,}9p^3q^2-9{,}9p^2q^3-2{,}5pq^4$$.