Упр.619 ГДЗ Никольский Потапов 7 класс (Алгебра)

а) (2a/(1-a))/(1-(1-a)/2a)^-1;
б) (2a/(2-a))/(2-(2-a)/2a)^-1;
в) (1/x — 1/(x+3))^-1 + (3/(x+3)-3/x)^-1;
г) (1/x — 1/(x-1))^-1 + (4/(x-1)-4/x)^-1.

1. а)

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

2. б)

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

3. в)

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

4. г)

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

Ответ

а) $$\frac{2a}{1-3a}$$; б) $$\frac{a}{2(1-a)}$$; в) $$\frac{2x^2+6x}{9}$$; г) $$\frac{3x-3x^2}{4}$$.