Упр.38.5 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) 3b/8 : b;
2) 6a/5b : (3a^2)/(20b^2 );
3) -9a/b^5 : (18a^4)/b^3 ;
4) a^2 : a/(b^2 c);
5) 36a/c^3 : (4a^2 c);
6) (16x^3 y^8)/(33z^5 ) : (-(12x^2)/(55z^6 )).

1. $$\frac{3b}{8}:b=\frac{3b}{8}\cdot\frac{1}{b}=\frac{3}{8}.$$
2. $$\frac{6a}{5b}:\frac{3a^2}{20b^2}=\frac{6a}{5b}\cdot\frac{20b^2}{3a^2}=\frac{6\cdot 20\cdot ab^2}{5\cdot 3\cdot ba^2}=\frac{8b}{a}.$$
3. $$-\frac{9a}{b^5}:\frac{18a^4}{b^3}=-\frac{9a}{b^5}\cdot\frac{b^3}{18a^4}=-\frac{9}{18}\cdot\frac{1}{a^3b^2}=-\frac{1}{2a^3b^2}.$$
4. $$a^2:\frac{a}{b^2c}=a^2\cdot\frac{b^2c}{a}=ab^2c.$$
5. $$\frac{36a}{c^3}:(4a^2c)=\frac{36a}{c^3}\cdot\frac{1}{4a^2c}=\frac{36}{4}\cdot\frac{1}{ac^4}=\frac{9}{ac^4}.$$
6. $$\frac{16x^3y^8}{33z^5}:\left(-\frac{12x^2}{55z^6}\right)=\frac{16x^3y^8}{33z^5}\cdot\left(-\frac{55z^6}{12x^2}\right).$$
   $$=-\frac{16\cdot 55}{33\cdot 12}\cdot xy^8z=-\frac{20xy^8z}{9}.$$

Ответ

1) $$\frac{3}{8}$$; 2) $$\frac{8b}{a}$$; 3) $$-\frac{1}{2a^3b^2}$$; 4) $$ab^2c$$; 5) $$\frac{9}{ac^4}$$; 6) $$-\frac{20xy^8z}{9}$$.