Упр.538 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1) a^5/(6c^3):a^2/(4c^3); 2) 9a^2/m^3:6a^2/m^5;
3) (4a/b^3)^2·b^4/(8a); 4) (3c/k^2)^3:9c/k^3;
5) 5a/(28b^2)·8ab·7b/(5a^3); 6) (-(25a^4 b^3)/(14c^2))·(-21c/(10a^3 b^3));
7) (4x(x-1)+1)/(4-x^2):((1-2x)/(x-2)); 8) (x^2-4(x-1))/(x-1):((2-x)/(1-x^2)).

1. $$\frac{a^5}{6c^3}:\frac{a^2}{4c^3}=\frac{a^5}{6c^3}\cdot\frac{4c^3}{a^2}=\frac{4a^3}{6}=\frac{2a^3}{3}.$$
2. $$\frac{9a^2}{m^3}:\frac{6a^2}{m^5}=\frac{9a^2}{m^3}\cdot\frac{m^5}{6a^2}=\frac{9m^2}{6}=\frac{3m^2}{2}.$$
3. $$\left(\frac{4a}{b^3}\right)^2\cdot\frac{b^4}{8a}=\frac{16a^2}{b^6}\cdot\frac{b^4}{8a}=\frac{2a}{b^2}.$$
4. $$\left(\frac{3c}{k^2}\right)^3:\frac{9c}{k^3}=\frac{27c^3}{k^6}\cdot\frac{k^3}{9c}=\frac{3c^2}{k^3}.$$
5. $$\frac{5a}{28b^2}\cdot 8ab\cdot\frac{7b}{5a^3}=\frac{5\cdot 8\cdot 7\cdot a\cdot a\cdot b\cdot b}{28\cdot 5\cdot a^3\cdot b^2}=\frac{8}{4a}=\frac{2}{a}.$$
6. $$\left(-\frac{25a^4b^3}{14c^2}\right)\cdot\left(-\frac{21c}{10a^3b^3}\right)=\frac{25\cdot 21}{14\cdot 10}\cdot\frac{a^4}{a^3}\cdot\frac{c}{c^2}=\frac{15a}{4c}.$$
7. $$\frac{4x(x-1)+1}{4-x^2}:\frac{1-2x}{x-2}=\frac{4x^2-4x+1}{4-x^2}\cdot\frac{x-2}{1-2x}.$$
   Так как
   $$4x^2-4x+1=(2x-1)^2,\quad 4-x^2=(2-x)(2+x),\quad 1-2x=-(2x-1),$$
   то
   $$\frac{(2x-1)^2}{(2-x)(2+x)}\cdot\frac{x-2}{1-2x}=\frac{2x-1}{2+x}.$$
8. $$\frac{x^2-4(x-1)}{x-1}:\frac{2-x}{1-x^2}=\frac{x^2-4x+4}{x-1}\cdot\frac{1-x^2}{2-x}.$$
   Далее
   $$x^2-4x+4=(x-2)^2,\quad 1-x^2=(1-x)(1+x)=-(x-1)(x+1),\quad 2-x=-(x-2).$$
   Тогда
   $$\frac{(x-2)^2}{x-1}\cdot\frac{1-x^2}{2-x}=(x-2)(x+1).$$

Ответ

1) $$\frac{2a^3}{3}$$; 2) $$\frac{3m^2}{2}$$; 3) $$\frac{2a}{b^2}$$; 4) $$\frac{3c^2}{k^3}$$; 5) $$\frac{2}{a}$$; 6) $$\frac{15a}{4c}$$; 7) $$\frac{2x-1}{x+2}$$; 8) $$\left(x-2\right)\left(x+1\right)$$.