Упр.15.3 ГДЗ Мордкович Семенов 8 класс (Алгебра)

а) (10y^5)/9a : (5y^3)/3y · (3a^3)/by;
б) (25a^3 b^3)/(14x^2 y^4) · (21xy^3)/(10a^2 b^2) · (8xy^2)/15ab;
в) (28a^2)/(27x^3) : (21x^4)/45y · x^8/20ya;
г) (45m^4)/(49n^2 t) · (56n^3)/(27m^2) : (20m^2 n)/(63t^2);
д) 35xy/(24a^2) : (42x^2 b)/5ay : 15ab/2x;
е) 30x/(23a^3) : (y^2 b)/46ax : 20ab/(3x^2).

1. а)

   $$\frac{10y^5}{9a}:\frac{5y^3}{3y}\cdot\frac{3a^3}{by} =\frac{10y^5}{9a}\cdot\frac{3y}{5y^3}\cdot\frac{3a^3}{by}$$
   $$=\frac{10\cdot 3\cdot 3\; a^3 y^6}{9\cdot 5\; a\cdot b\cdot y^4} =\frac{2a^2y^2}{b}$$

2. б)

   $$\frac{25a^3b^3}{14x^2y^4}\cdot\frac{21xy^3}{10a^2b^2}\cdot\frac{8xy^2}{15ab}$$
   $$=\frac{25\cdot 21\cdot 8\; a^3b^3xy^5}{14\cdot 10\cdot 15\; a^3b^3x^2y^4} =\frac{2y}{1}=2y$$

3. в)

   $$\frac{28a^2}{27x^3}:\frac{21x^4}{45y}\cdot\frac{x^8}{20ya} =\frac{28a^2}{27x^3}\cdot\frac{45y}{21x^4}\cdot\frac{x^8}{20ya}$$
   $$=\frac{28\cdot 45}{27\cdot 21\cdot 20}\cdot\frac{a^2x^8}{a x^7} =\frac{a x}{9}$$

4. г)

   $$\frac{45m^4}{49n^2t}\cdot\frac{56n^3}{27m^2}:\frac{20m^2n}{63t^2} =\frac{45m^4}{49n^2t}\cdot\frac{56n^3}{27m^2}\cdot\frac{63t^2}{20m^2n}$$
   $$=\frac{45\cdot 56\cdot 63}{49\cdot 27\cdot 20}\cdot\frac{m^4n^3t^2}{m^4n^3t} =6t$$

5. д)

   $$\frac{35xy}{24a^2}:\frac{42x^2b}{5ay}:\frac{15ab}{2x} =\frac{35xy}{24a^2}\cdot\frac{5ay}{42x^2b}\cdot\frac{2x}{15ab}$$
   $$=\frac{35\cdot 5\cdot 2}{24\cdot 42\cdot 15}\cdot\frac{x y^2}{a^2 b^2 x} =\frac{5y^2}{216a^2b^2}$$

6. е)

   $$\frac{30x}{23a^3}:\frac{y^2b}{46ax}:\frac{20ab}{3x^2} =\frac{30x}{23a^3}\cdot\frac{46ax}{y^2b}\cdot\frac{3x^2}{20ab}$$
   $$=\frac{30\cdot 46\cdot 3}{23\cdot 20}\cdot\frac{x^4}{a^3b\cdot a b\cdot y^2} =\frac{9x^4}{a^3b^2y^2}$$

Ответ

а) $$\frac{2a^2y^2}{b}$$; б) $$2y$$; в) $$\frac{ax}{9}$$; г) $$6t$$; д) $$\frac{5y^2}{216a^2b^2}$$; е) $$\frac{9x^4}{a^3b^2y^2}$$.