Упр.16.2 ГДЗ Мордкович Семенов 8 класс (Алгебра)

1. а)

   $$\frac{p-5}{p+6}:\left(p-\frac{11p}{p+6}\right)=\frac{p-5}{p+6}:\frac{p(p+6)-11p}{p+6}$$
   $$=\frac{p-5}{p+6}:\frac{p^2-5p}{p+6}=\frac{p-5}{p+6}\cdot\frac{p+6}{p(p-5)}=\frac{1}{p}.$$

2. б)

   $$\left(\frac{q}{q-5}-2q\right):\frac{11-2q}{q-5}=\frac{q-2q(q-5)}{q-5}:\frac{11-2q}{q-5}$$
   $$=\frac{q-2q^2+10q}{q-5}:\frac{11-2q}{q-5}=\frac{11q-2q^2}{q-5}:\frac{11-2q}{q-5}$$
   $$=\frac{q(11-2q)}{q-5}\cdot\frac{q-5}{11-2q}=q.$$

3. в)

   $$\frac{cd-d^2}{c^2+d^2}\cdot\left(\frac{c}{c+d}+\frac{d}{c-d}\right)=\frac{cd-d^2}{c^2+d^2}\cdot\frac{c(c-d)+d(c+d)}{(c-d)(c+d)}$$
   $$=\frac{cd-d^2}{c^2+d^2}\cdot\frac{c^2-dc+dc+d^2}{(c-d)(c+d)}=\frac{cd-d^2}{c^2+d^2}\cdot\frac{c^2+d^2}{(c-d)(c+d)}$$
   $$=\frac{d(c-d)}{(c-d)(c+d)}=\frac{d}{c+d}.$$

4. г)

   $$\left(\frac{6}{x-y}-\frac{5}{x+y}\right)\cdot\frac{x-y}{x+11y}=\frac{6(x+y)-5(x-y)}{(x-y)(x+y)}\cdot\frac{x-y}{x+11y}$$
   $$=\frac{6x+6y-5x+5y}{(x-y)(x+y)}\cdot\frac{x-y}{x+11y}=\frac{x+11y}{(x-y)(x+y)}\cdot\frac{x-y}{x+11y}=\frac{1}{x+y}.$$

5. д)

   $$\frac{x^2-9}{2x^2+1}\cdot\left(\frac{6x+1}{x-3}+\frac{6x-1}{x+3}\right)=\frac{x^2-9}{2x^2+1}\cdot\frac{(6x+1)(x+3)+(6x-1)(x-3)}{(x-3)(x+3)}$$
   $$=\frac{x^2-9}{2x^2+1}\cdot\frac{6x^2+18x+x+3+6x^2-18x-x+3}{(x-3)(x+3)}$$
   $$=\frac{x^2-9}{2x^2+1}\cdot\frac{12x^2+6}{(x-3)(x+3)}=\frac{x^2-9}{2x^2+1}\cdot\frac{6(2x^2+1)}{x^2-9}=6.$$

6. е)

   $$\frac{b+3}{b^3+9b}\cdot\left(\frac{b+3}{b-3}+\frac{b-3}{b+3}\right)=\frac{b+3}{b(b^2+9)}\cdot\frac{(b+3)^2+(b-3)^2}{(b-3)(b+3)}$$
   $$=\frac{b+3}{b(b^2+9)}\cdot\frac{b^2+6b+9+b^2-6b+9}{(b-3)(b+3)}$$
   $$=\frac{b+3}{b(b^2+9)}\cdot\frac{2b^2+18}{(b-3)(b+3)}=\frac{b+3}{b(b^2+9)}\cdot\frac{2(b^2+9)}{(b-3)(b+3)}$$
   $$=\frac{2}{b(b-3)}=\frac{2}{b^2-3b}.$$

Ответ

а) $$\frac{1}{p}$$; б) $$q$$; в) $$\frac{d}{c+d}$$; г) $$\frac{1}{x+y}$$; д) $$6$$; е) $$\frac{2}{b^2-3b}$$.