Упр.538 ГДЗ Никольский Потапов 7 класс (Алгебра)

б) ((c+3)/(c-3)-c/(c+3))*(c-3)/(c+1);
в) ((14+a2)/(a2-4)-(a-4)/(a+2))*(a-2)/6;
г) (a/(a-4)-(a-4)/(a+4))*(a+4)/4;
д) ((y+1)/(y-1)-(y-1)/(y+1))*(y+1)/4y;
е) ((1+a)/(1-a)-(1-a)/(1+a)):2a/(1-a);
ж) 4y/(y-1) * (y/8-1/4+1/8y);
з) (a/8+1/3+1/6a):(a+1)/12a.

1. а)

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

2. б)

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

3. в)

   $$\left(\frac{14+a^2}{a^2-4}-\frac{a-4}{a+2}\right)\cdot\frac{a-2}{6} =\left(\frac{14+a^2}{(a-2)(a+2)}-\frac{(a-4)(a-2)}{(a+2)(a-2)}\right)\cdot\frac{a-2}{6}$$
   $$=\frac{14+a^2-(a^2-2a-4a+8)}{(a-2)(a+2)}\cdot\frac{a-2}{6} =\frac{6a+6}{(a-2)(a+2)}\cdot\frac{a-2}{6}$$
   $$=\frac{6(a+1)}{(a-2)(a+2)}\cdot\frac{a-2}{6} =\frac{a+1}{a+2}.$$

4. г)

   $$\left(\frac{a}{a-4}-\frac{a-4}{a+4}\right)\cdot\frac{a+4}{4} =\frac{a(a+4)-(a-4)^2}{(a-4)(a+4)}\cdot\frac{a+4}{4}$$
   $$=\frac{a^2+4a-(a^2-8a+16)}{(a-4)(a+4)}\cdot\frac{a+4}{4} =\frac{12a-16}{(a-4)(a+4)}\cdot\frac{a+4}{4}$$
   $$=\frac{4(3a-4)}{4(a-4)} =\frac{3a-4}{a-4}.$$

5. д)

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

6. е)

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

7. ж)

   $$\frac{4y}{y-1}\left(\frac{y}{8}-\frac14+\frac{1}{8y}\right) =\frac{4y}{y-1}\cdot\frac{y^2-2y+1}{8y}$$
   $$=\frac{4y}{y-1}\cdot\frac{(y-1)^2}{8y} =\frac{y-1}{2}.$$

8. з)

   $$\left(\frac{a}{8}+\frac13+\frac{1}{6a}\right):\frac{a+1}{12a} =\frac{3a^2+8a+4}{24a}:\frac{a+1}{12a}$$
   $$=\frac{3a^2+8a+4}{24a}\cdot\frac{12a}{a+1} =\frac{3a^2+8a+4}{2(a+1)}.$$

Ответ

а) $$\frac{1}{a}$$; б) $$\frac{9}{c+3}$$; в) $$\frac{a+1}{a+2}$$; г) $$\frac{3a-4}{a-4}$$; д) $$\frac{1}{y-1}$$; е) $$\frac{2}{1+a}$$; ж) $$\frac{y-1}{2}$$; з) $$\frac{3a^2+8a+4}{2(a+1)}$$.