Упр.38.16 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) (4c-d)/(c^2+cd)·(2c^2-2d^2)/(4c^2-cd);
2) (b^2-6b+9)/(b^2-3b+9)·(b^3+27)/(5b-15);
3) (a^3-16a)/(3a^2 b)·(12ab^2)/(4a+16);
4) (a^3+b^3)/(a^2-b^2 )·(7a-7b)/(a^2-ab+b^2 );
5) (m+2n)/(2-3m) : (m^2+4mn+4n^2)/(3m^2-2m);
6) (a^3+8)/(16-a^4 ) : (a^2-2a+4)/(a^2+4);
7) (x^2-12x+36)/(3x+21)·(x^2-49)/(4x-24);
8) (3a+15b)/(a^2-81b^2 ) : (4a+20b)/(a^2-18ab+81b^2 ).

1. $$\frac{4c-d}{c^2+cd}\cdot\frac{2c^2-2d^2}{4c^2-cd}= \frac{4c-d}{c(c+d)}\cdot\frac{2(c-d)(c+d)}{c(4c-d)}$$

   Сократим одинаковые множители:
   $$=\frac{2(c-d)}{c^2}$$

2. $$\frac{b^2-6b+9}{b^2-3b+9}\cdot\frac{b^3+27}{5b-15}= \frac{(b-3)^2}{b^2-3b+9}\cdot\frac{(b+3)(b^2-3b+9)}{5(b-3)}$$

   $$=\frac{(b-3)(b+3)}{5}=\frac{b^2-9}{5}$$

3. $$\frac{a^3-16a}{3a^2b}\cdot\frac{12ab^2}{4a+16}= \frac{a(a^2-16)}{3a^2b}\cdot\frac{12ab^2}{4(a+4)}$$

   $$=\frac{a(a-4)(a+4)}{3a^2b}\cdot\frac{12ab^2}{4(a+4)}=b(a-4)$$

4. $$\frac{a^3+b^3}{a^2-b^2}\cdot\frac{7a-7b}{a^2-ab+b^2}= \frac{(a+b)(a^2-ab+b^2)}{(a-b)(a+b)}\cdot\frac{7(a-b)}{a^2-ab+b^2}$$

   $$=7$$

5. $$\frac{m+2n}{2-3m}:\frac{m^2+4mn+4n^2}{3m^2-2m}= \frac{m+2n}{2-3m}\cdot\frac{3m^2-2m}{(m+2n)^2}$$

   $$=\frac{m+2n}{2-3m}\cdot\frac{m(3m-2)}{(m+2n)^2} =-\frac{m}{m+2n}$$

6. $$\frac{a^3+8}{16-a^4}:\frac{a^2-2a+4}{a^2+4}= \frac{(a+2)(a^2-2a+4)}{(4-a^2)(a^2+4)}\cdot\frac{a^2+4}{a^2-2a+4}$$

   $$=\frac{a+2}{4-a^2}=\frac{a+2}{(2-a)(2+a)}=\frac{1}{2-a}$$

7. $$\frac{x^2-12x+36}{3x+21}\cdot\frac{x^2-49}{4x-24}= \frac{(x-6)^2}{3(x+7)}\cdot\frac{(x-7)(x+7)}{4(x-6)}$$

   $$=\frac{(x-6)(x-7)}{12}$$

8. $$\frac{3a+15b}{a^2-81b^2}:\frac{4a+20b}{a^2-18ab+81b^2}= \frac{3(a+5b)}{(a-9b)(a+9b)}:\frac{4(a+5b)}{(a-9b)^2}$$

   $$=\frac{3(a+5b)}{(a-9b)(a+9b)}\cdot\frac{(a-9b)^2}{4(a+5b)} =\frac{3(a-9b)}{4(a+9b)}$$

Ответ

1. $$\frac{2(c-d)}{c^2}$$
2. $$\frac{b^2-9}{5}$$
3. $$b(a-4)$$
4. $$7$$
5. $$-\frac{m}{m+2n}$$
6. $$\frac{1}{2-a}$$
7. $$\frac{(x-6)(x-7)}{12}$$
8. $$\frac{3(a-9b)}{4(a+9b)}$$