Упр.535 ГДЗ Никольский Потапов 7 класс (Алгебра)

б) (a/(a-1)+1):(1-a/(a-1));
в) (m-1/(1+m))*(m+1)/(1-m-m2);
г) (a+a2/c):(b+bc/a);
д) ((a+x)/x — 2x/(x-a)):(a2+x2)/(x-a);
е) ((x2+1)/(2x-1) — x/2)*(1-2x)/(x+2);
ж) (n/(n+x) — 2x/(x-a)):(a2+x2)/(x-a);
з) 3/5x-3/(x+y)*((x+y)/5x-x-y).

1. а)

   $$\left(\frac{a+x}{a}-\frac{x-y}{x}\right)\cdot \frac{a^2}{x^2+ay} =\frac{x(a+x)-a(x-y)}{ax}\cdot \frac{a^2}{x^2+ay}$$
   $$=\frac{ax+x^2-ax+ay}{ax}\cdot \frac{a^2}{x^2+ay} =\frac{x^2+ay}{ax}\cdot \frac{a^2}{x^2+ay} =\frac{a}{x}.$$

2. б)

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

3. в)

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

4. г)

   $$\left(a+\frac{a^2}{c}\right):\left(b+\frac{bc}{a}\right) =\frac{ac+a^2}{c}:\frac{ab+bc}{a}$$
   $$=\frac{a(c+a)}{c}:\frac{b(a+c)}{a} =\frac{a(c+a)}{c}\cdot \frac{a}{b(a+c)} =\frac{a^2}{bc}.$$

5. д)

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

6. е)

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

7. ж)

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

8. з)

   $$\frac{3}{5x}-\frac{3}{x+y}\cdot \left(\frac{x+y}{5x}-x-y\right)$$
   $$=\frac{3}{5x}-\frac{3}{x+y}\cdot \frac{x+y-5x^2-5xy}{5x}$$
   $$=\frac{3}{5x}-\frac{3(x+y-5x^2-5xy)}{5x(x+y)}$$
   $$=\frac{3(x+y)-3(x+y-5x^2-5xy)}{5x(x+y)}$$
   $$=\frac{15x^2+15xy}{5x(x+y)}=3.$$

Ответ

а) $$\frac{a}{x}$$; б) $$1-2a$$; в) $$-1$$; г) $$\frac{a^2}{bc}$$; д) $$-\frac{1}{x}$$; е) $$-\frac{1}{2}$$; ж) $$-\frac{x}{n}$$; з) $$3$$.