Упр.35 Вариант 2 Дидактические материалы ГДЗ Мерзляк Полонский 8 класс (Алгебра)

1) ((a+3)/(a-3)+(a-3)/(a+3)) :(3a^2+27)/(9-a^2);
2) (5x-10x/(x+1)) :(15x-15)/(4x+4);
3) 3a/(a-4)-(a+2)/(5a-20)•240/(a^2+2a);
4) (8b/(b+7)-15b/(b^2+14b+49)) :(8b+41)/(b^2-49)+(7b-49)/(b+7);
5) ((a-b)/(a^2+ab)-a/(ab+b^2 )) :(b^2/(a^3-ab^2 )+1/(a+b));
6) (x^2+5x)/(x-5)^2 :(5/(x+5)+(x^2+25)/(x^2-25)-5/(5-x)).

1. $$\left(\frac{a+3}{a-3}+\frac{a-3}{a+3}\right):\frac{3a^2+27}{9-a^2}$$
   $$=\frac{(a+3)^2+(a-3)^2}{(a-3)(a+3)}\cdot\frac{9-a^2}{3a^2+27}$$
   $$=\frac{a^2+6a+9+a^2-6a+9}{a^2-9}\cdot\frac{9-a^2}{3(a^2+9)}$$
   $$=\frac{2(a^2+9)}{a^2-9}\cdot\frac{9-a^2}{3(a^2+9)}=-\frac23.$$

2. $$\left(5x-\frac{10x}{x+1}\right):\frac{15x-15}{4x+4}$$
   $$=\frac{5x(x+1)-10x}{x+1}\cdot\frac{4x+4}{15x-15}$$
   $$=\frac{5x^2-5x}{x+1}\cdot\frac{4(x+1)}{15(x-1)}$$
   $$=\frac{5x(x-1)\cdot 4}{15(x-1)}=\frac{4x}{3}.$$

3. $$\frac{3a}{a-4}-\frac{a+2}{5a-20}\cdot\frac{240}{a^2+2a}$$
   $$=\frac{3a}{a-4}-\frac{a+2}{5(a-4)}\cdot\frac{240}{a(a+2)}$$
   $$=\frac{3a}{a-4}-\frac{48}{a(a-4)}$$
   $$=\frac{3a^2-48}{a(a-4)}=\frac{3(a^2-16)}{a(a-4)}$$
   $$=\frac{3(a-4)(a+4)}{a(a-4)}=\frac{3(a+4)}{a}.$$

4. $$\left(\frac{8b}{b+7}-\frac{15b}{b^2+14b+49}\right):\frac{8b+41}{b^2-49}+\frac{7b-49}{b+7}$$
   $$=\left(\frac{8b}{b+7}-\frac{15b}{(b+7)^2}\right)\cdot\frac{b^2-49}{8b+41}+\frac{7b-49}{b+7}$$
   $$=\frac{8b(b+7)-15b}{(b+7)^2}\cdot\frac{(b-7)(b+7)}{8b+41}+\frac{7(b-7)}{b+7}$$
   $$=\frac{b(8b+41)(b-7)}{(b+7)(8b+41)}+\frac{7(b-7)}{b+7}$$
   $$=\frac{b(b-7)}{b+7}+\frac{7(b-7)}{b+7}=\frac{(b-7)(b+7)}{b+7}=b-7.$$

5. $$\left(\frac{a-b}{a^2+ab}-\frac{a}{ab+b^2}\right):\left(\frac{b^2}{a^3-ab^2}+\frac{1}{a+b}\right)$$
   $$=\left(\frac{a-b}{a(a+b)}-\frac{a}{b(a+b)}\right):\left(\frac{b^2}{a(a-b)(a+b)}+\frac{1}{a+b}\right)$$
   $$=\frac{b(a-b)-a^2}{ab(a+b)}:\frac{b^2+a(a-b)}{a(a-b)(a+b)}$$
   $$=\frac{ab-b^2-a^2}{ab(a+b)}\cdot\frac{a(a-b)(a+b)}{a^2-ab+b^2}$$
   $$=-\frac{a-b}{b}=\frac{b-a}{b}.$$

6. $$\frac{x^2+5x}{(x-5)^2}:\left(\frac{5}{x+5}+\frac{x^2+25}{x^2-25}-\frac{5}{5-x}\right)$$
   $$=\frac{x(x+5)}{(x-5)^2}:\left(\frac{5}{x+5}+\frac{x^2+25}{(x-5)(x+5)}+\frac{5}{x-5}\right)$$
   $$=\frac{x(x+5)}{(x-5)^2}:\frac{5(x-5)+x^2+25+5(x+5)}{(x-5)(x+5)}$$
   $$=\frac{x(x+5)}{(x-5)^2}:\frac{x^2+10x+25}{(x-5)(x+5)}$$
   $$=\frac{x(x+5)}{(x-5)^2}\cdot\frac{(x-5)(x+5)}{(x+5)^2}=\frac{x}{x-5}.$$

Ответ

1) $$-\frac23$$; 2) $$\frac{4x}{3}$$; 3) $$\frac{3(a+4)}{a}$$; 4) $$b-7$$; 5) $$\frac{b-a}{b}$$; 6) $$\frac{x}{x-5}$$.