Упр.39.4 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) (x^2+14x+49)/(x+6) : (13/(x+6)-x+6);
2) (c-(2c-9)/(c+8)) : (c^2+3c)/(c^2-64)+24/c;
3) (36/(x^2-9)-(x-3)/(x+3)-(3+x)/(3-x)) : 6/(3-x);
4) ((2y-1)/(y^2+2y+4)+(9y+6)/(y^3-8)+1/(y-2))·(y^2-4)/18.

1. $$\frac{x^2+14x+49}{x+6}:\left(\frac{13}{x+6}-x+6\right)$$
   $$=\frac{(x+7)^2}{x+6}:\left(\frac{13-(x-6)(x+6)}{x+6}\right)$$
   $$=\frac{(x+7)^2}{x+6}:\frac{13-x^2+36}{x+6}$$
   $$=\frac{(x+7)^2}{x+6}:\frac{49-x^2}{x+6}$$
   $$=\frac{(x+7)^2}{x+6}\cdot\frac{x+6}{(7-x)(7+x)}$$
   $$=\frac{x+7}{7-x}.$$
2. $$\left(c-\frac{2c-9}{c+8}\right):\frac{c^2+3c}{c^2-64}+\frac{24}{c}$$
   $$=\frac{c(c+8)-(2c-9)}{c+8}:\frac{c(c+3)}{(c-8)(c+8)}+\frac{24}{c}$$
   $$=\frac{c^2+8c-2c+9}{c+8}\cdot\frac{(c-8)(c+8)}{c(c+3)}+\frac{24}{c}$$
   $$=\frac{c^2+6c+9}{c+8}\cdot\frac{(c-8)(c+8)}{c(c+3)}+\frac{24}{c}$$
   $$=\frac{(c+3)^2(c-8)}{c(c+3)}+\frac{24}{c}$$
   $$=\frac{(c+3)(c-8)}{c}+\frac{24}{c}$$
   $$=\frac{c^2-5c-24+24}{c}=\frac{c^2-5c}{c}=c-5.$$
3. $$\left(\frac{36}{x^2-9}-\frac{x-3}{x+3}-\frac{3+x}{3-x}\right):\frac{6}{3-x}$$
   $$=\left(\frac{36}{(x-3)(x+3)}-\frac{x-3}{x+3}+\frac{x+3}{x-3}\right)\cdot\frac{3-x}{6}$$
   $$=\frac{36-(x-3)^2+(x+3)^2}{(x-3)(x+3)}\cdot\frac{3-x}{6}$$
   $$=\frac{36-(x^2-6x+9)+(x^2+6x+9)}{(x-3)(x+3)}\cdot\frac{3-x}{6}$$
   $$=\frac{12x+36}{(x-3)(x+3)}\cdot\frac{3-x}{6}$$
   $$=\frac{12(x+3)(3-x)}{(x-3)(x+3)\cdot 6}=-2.$$
4. $$\left(\frac{2y-1}{y^2+2y+4}+\frac{9y+6}{y^3-8}+\frac{1}{y-2}\right)\cdot\frac{y^2-4}{18}$$
   $$=\left(\frac{2y-1}{y^2+2y+4}+\frac{9y+6}{(y-2)(y^2+2y+4)}+\frac{y^2+2y+4}{(y-2)(y^2+2y+4)}\right)\cdot\frac{y^2-4}{18}$$
   $$=\frac{(2y-1)(y-2)+9y+6+y^2+2y+4}{(y-2)(y^2+2y+4)}\cdot\frac{y^2-4}{18}$$
   $$=\frac{2y^2-5y+2+9y+6+y^2+2y+4}{(y-2)(y^2+2y+4)}\cdot\frac{y^2-4}{18}$$
   $$=\frac{3y^2+6y+12}{(y-2)(y^2+2y+4)}\cdot\frac{y^2-4}{18}$$
   $$=\frac{3(y^2+2y+4)}{(y-2)(y^2+2y+4)}\cdot\frac{(y-2)(y+2)}{18}$$
   $$=\frac{y+2}{6}.$$

Ответ

1) $$\frac{x+7}{7-x}$$; 2) $$c-5$$; 3) $$-2$$; 4) $$\frac{y+2}{6}$$.