Упр.2.45 ГДЗ Мордкович 8 класс (Алгебра)

2.45

1. а)

   $$3n^2-3m^2=3(n-m)(n+m),$$
   $$m^2-2mn+n^2=(m-n)^2,$$
   $$m^2+2mn+n^2=(m+n)^2.$$

   Общий знаменатель:
   $$3(m-n)^2(m+n)^2.$$

   Тогда
   $$\frac{2mn}{3n^2-3m^2}=\frac{2mn(m^2-n^2)}{3(m-n)^2(m+n)^2},$$
   $$\frac{m^2}{m^2-2mn+n^2}=\frac{3m^2(m+n)^2}{3(m-n)^2(m+n)^2},$$
   $$\frac{n^2}{m^2+2mn+n^2}=\frac{3n^2(m-n)^2}{3(m-n)^2(m+n)^2}.$$

2. б)

   $$3n^2-3m^2=3(n-m)(n+m),$$
   $$-m^2+2mn-n^2=-(m-n)^2,$$
   $$2mn+m^2+n^2=(m+n)^2.$$

   Общий знаменатель:
   $$3(m-n)^2(m+n)^2.$$

   Тогда
   $$\frac{2mn}{3n^2-3m^2}=\frac{-2mn(m^2-n^2)}{3(m-n)^2(m+n)^2},$$
   $$\frac{(m+n)^2}{-m^2+2mn-n^2}=\frac{-3(m+n)^4}{3(m-n)^2(m+n)^2},$$
   $$\frac{(m-n)^2}{2mn+m^2+n^2}=\frac{3(m-n)^4}{3(m-n)^2(m+n)^2}.$$

3. в)

   $$2y^2-2x^2=2(y-x)(y+x),$$
   $$x^2+2xy+y^2=(x+y)^2,$$
   $$x^2-2xy+y^2=(x-y)^2.$$

   Общий знаменатель:
   $$2(x-y)^2(x+y)^2.$$

   Тогда
   $$\frac{5xy}{2y^2-2x^2}=\frac{5xy(y^2-x^2)}{2(x-y)^2(x+y)^2},$$
   $$\frac{x^2}{x^2+2xy+y^2}=\frac{2x^2(x-y)^2}{2(x-y)^2(x+y)^2},$$
   $$\frac{3y^2}{x^2-2xy+y^2}=\frac{6y^2(x+y)^2}{2(x-y)^2(x+y)^2}.$$

4. г)

   $$5x^2-45=5(x-3)(x+3),$$
   $$-x^2-6x-9=-(x+3)^2,$$
   $$x^2+9-6x=(x-3)^2.$$

   Общий знаменатель:
   $$5(x-3)^2(x+3)^2.$$

   Тогда
   $$\frac{6x}{5x^2-45}=\frac{6x(x-3)(x+3)}{5(x-3)^2(x+3)^2}=\frac{6x(x^2-9)}{5(x-3)^2(x+3)^2},$$
   $$\frac{(x-3)^2}{-x^2-6x-9}=\frac{-5(x-3)^4}{5(x-3)^2(x+3)^2},$$
   $$\frac{x^2+6x+9}{x^2+9-6x}=\frac{5(x+3)^4}{5(x-3)^2(x+3)^2}.$$

Ответ

а) $$\frac{2mn(m^2-n^2)}{3(m-n)^2(m+n)^2},\ \frac{3m^2(m+n)^2}{3(m-n)^2(m+n)^2},\ \frac{3n^2(m-n)^2}{3(m-n)^2(m+n)^2}$$

б) $$\frac{-2mn(m^2-n^2)}{3(m-n)^2(m+n)^2},\ \frac{-3(m+n)^4}{3(m-n)^2(m+n)^2},\ \frac{3(m-n)^4}{3(m-n)^2(m+n)^2}$$

в) $$\frac{5xy(y^2-x^2)}{2(x-y)^2(x+y)^2},\ \frac{2x^2(x-y)^2}{2(x-y)^2(x+y)^2},\ \frac{6y^2(x+y)^2}{2(x-y)^2(x+y)^2}$$

г) $$\frac{6x(x^2-9)}{5(x-3)^2(x+3)^2},\ \frac{-5(x-3)^4}{5(x-3)^2(x+3)^2},\ \frac{5(x+3)^4}{5(x-3)^2(x+3)^2}$$