Упр.716 ГДЗ Никольский Потапов 8 класс (Алгебра)

Рассмотрим вариант решения задания из учебника Никольский, Потапов 8 класс, Просвещение: 716

1. а)

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

2. б)

   $$\left(a^2-\frac{1+a^4}{a^2+1}\right):\frac{1-a}{1+a^2} =\frac{a^2(a^2+1)-1-a^4}{a^2+1}\cdot\frac{1+a^2}{1-a}$$
   $$=\frac{a^4+a^2-1-a^4}{1}\cdot\frac{1}{1-a} =\frac{a^2-1}{1-a} =\frac{(a-1)(a+1)}{-(a-1)} =-a-1.$$

3. в)

   $$\left(\frac{1}{m^2-m}-\frac{1}{m-1}\right)\cdot\frac{1}{m+2}+\frac{m}{m^2-4}$$
   $$=\left(\frac{1}{m(m-1)}-\frac{1}{m-1}\right)\cdot\frac{1}{m+2}+\frac{m}{(m-2)(m+2)}$$
   $$=\frac{1-m}{m(m-1)}\cdot\frac{1}{m+2}+\frac{m}{(m-2)(m+2)} =-\frac{1}{m(m+2)}+\frac{m}{(m-2)(m+2)}$$
   $$=\frac{-\,(m-2)+m^2}{m(m-2)(m+2)} =\frac{m^2-m+2}{m(m^2-4)}.$$

4. г)

   $$\left(\frac{k+4}{3k+3}-\frac{1}{k+1}\right)\cdot\frac{3}{k+1}-\frac{2}{1-k^2}$$
   $$=\left(\frac{k+4}{3(k+1)}-\frac{1}{k+1}\right)\cdot\frac{3}{k+1}+\frac{2}{k^2-1}$$
   $$=\frac{k+4-3}{3(k+1)}\cdot\frac{3}{k+1}+\frac{2}{(k-1)(k+1)} =\frac{k+1}{(k+1)^2}+\frac{2}{(k-1)(k+1)}$$
   $$=\frac{1}{k+1}+\frac{2}{(k-1)(k+1)} =\frac{k-1+2}{(k+1)(k-1)} =\frac{k+1}{(k+1)(k-1)} =\frac{1}{k-1}.$$

5. д)

   $$\frac{2c}{c^2-4}-\frac{1}{c-2}:\left(\frac{c+1}{2c-2}-\frac{1}{c-1}\right)$$
   $$=\frac{2c}{(c-2)(c+2)}-\frac{1}{c-2}:\left(\frac{c+1}{2(c-1)}-\frac{1}{c-1}\right)$$
   $$=\frac{2c}{(c-2)(c+2)}-\frac{1}{c-2}:\frac{c+1-2}{2(c-1)}$$
   $$=\frac{2c}{(c-2)(c+2)}-\frac{1}{c-2}:\frac{c-1}{2(c-1)} =\frac{2c}{(c-2)(c+2)}-\frac{2}{c-2}$$
   $$=\frac{2c-2(c+2)}{(c-2)(c+2)} =\frac{-4}{c^2-4} =\frac{4}{4-c^2}.$$

6. е)

   $$\frac{y^2}{y^2-1}+\frac{1}{y+1}:\left(\frac{1}{2-y}-\frac{2}{2y-y^2}\right)$$
   $$=\frac{y^2}{y^2-1}+\frac{1}{y+1}:\left(\frac{1}{2-y}-\frac{2}{y(2-y)}\right)$$
   $$=\frac{y^2}{y^2-1}+\frac{1}{y+1}:\frac{y-2}{y(2-y)}$$
   $$=\frac{y^2}{(y-1)(y+1)}+\frac{1}{y+1}\cdot\frac{y(2-y)}{-(2-y)}$$
   $$=\frac{y^2}{(y-1)(y+1)}-\frac{y}{y+1} =\frac{y^2-y(y-1)}{(y-1)(y+1)} =\frac{y}{y^2-1}.$$

7. ж)

   $$\frac{5p+6}{p^2-4}-\frac{p}{p^2-4}:\frac{p}{p-2}-\frac{p+2}{p-2}$$
   $$=\frac{5p+6}{p^2-4}-\frac{p}{(p-2)(p+2)}\cdot\frac{p-2}{p}-\frac{p+2}{p-2}$$
   $$=\frac{5p+6}{(p-2)(p+2)}-\frac{1}{p+2}-\frac{p+2}{p-2}$$
   $$=\frac{5p+6-(p-2)-(p+2)(p+2)}{(p-2)(p+2)}$$
   $$=\frac{5p+6-p+2-p^2-4p-4}{p^2-4} =\frac{-p^2+4}{p^2-4} =-1.$$

8. з)

   $$\frac{21-5a}{a^2-9}-\frac{a}{a^2-9}:\frac{a}{a+3}-\frac{a-3}{a+3}$$
   $$=\frac{21-5a}{a^2-9}-\frac{a}{(a-3)(a+3)}\cdot\frac{a+3}{a}-\frac{a-3}{a+3}$$
   $$=\frac{21-5a}{(a-3)(a+3)}-\frac{1}{a-3}-\frac{a-3}{a+3}$$
   $$=\frac{21-5a-(a+3)-(a-3)^2}{(a-3)(a+3)}$$
   $$=\frac{21-5a-a-3-a^2+6a-9}{a^2-9} =\frac{-a^2+9}{a^2-9} =-1.$$

Ответ

а) $$\frac{1}{1-x}$$; б) $$-a-1$$; в) $$\frac{m^2-m+2}{m(m^2-4)}$$; г) $$\frac{1}{k-1}$$; д) $$\frac{4}{4-c^2}$$; е) $$\frac{y}{y^2-1}$$; ж) $$-1$$; з) $$-1$$.