Упр.12.16 ГДЗ Мордкович Семенов 8 класс (Алгебра)

а) (2p — q)/(4p + 2q) и pq/(4p^2 — q^2);
б) (3x^2)/(x^2 — 9y^2) и (x + 3y)/(2x — 6y);
в) 5y/(y^2 — 100) и (y^2 — 10y + 100)/(y^2 — 10y);
г) (t^2 + 6t + 36)/(t^3 + 6t^2) и 3t/(t^2 — 36);
д) (x + 2)/(x^2 — 9) и (x + 3)/(x^2 — 6x + 9);
е) (a — 5)/(a^2 + 10a + 25) и (a — 1)/(a^2 — 25).
Приведите дроби к наименьшему общему знаменателю:
а) 7mn/((n — m)(n + m)), n^2/(n + m), m^2/(n — m);
б) (a — 2)/((a — 3)(a + 3)), a^2/(a + 3), 3/(a — 3);
в) 6b/(25 — b^2), (b — 5)/(b + 5), (b + 5)/(b — 5);
г) (a + x)/(a — x), 2ax/(a^2 — x^2), (a — x)/(a + x);
д) 1/(a — 4)^2, (a^2 + 16)/(a^2 — 16), 1/(a + 4)^2;
е) (y + 1)/(1 — y)^2, (1 — y)/(1 + y)^2, (y^2 + 1)/(y^2 — 1).

1. а) $$4p+2q=2(2p+q), \qquad 4p^2-q^2=(2p-q)(2p+q).$$

   НОЗ: $$2(2p-q)(2p+q).$$

   $$\frac{2p-q}{4p+2q}=\frac{2p-q}{2(2p+q)}=\frac{(2p-q)^2}{2(2p-q)(2p+q)}$$
   $$\frac{pq}{4p^2-q^2}=\frac{2pq}{2(2p-q)(2p+q)}$$

2. б) $$x^2-9y^2=(x-3y)(x+3y), \qquad 2x-6y=2(x-3y).$$

   НОЗ: $$2(x-3y)(x+3y)=2(x^2-9y^2).$$

   $$\frac{3x^2}{x^2-9y^2}=\frac{6x^2}{2(x^2-9y^2)}$$
   $$\frac{x+3y}{2x-6y}=\frac{(x+3y)^2}{2(x^2-9y^2)}$$

3. в) $$y^2-100=(y-10)(y+10), \qquad y^2-10y=y(y-10).$$

   НОЗ: $$y(y-10)(y+10)=y(y^2-100).$$

   $$\frac{5y}{y^2-100}=\frac{5y^2}{y(y^2-100)}$$
   $$\frac{y^2-10y+100}{y^2-10y}=\frac{(y^2-10y+100)(y+10)}{y(y^2-100)}=\frac{y^3+1000}{y(y^2-100)}$$

4. г) $$t^3+6t^2=t^2(t+6), \qquad t^2-36=(t-6)(t+6).$$

   НОЗ: $$t^2(t-6)(t+6)=t^2(t^2-36).$$

   $$\frac{t^2+6t+36}{t^3+6t^2}=\frac{(t^2+6t+36)(t-6)}{t^2(t+6)(t-6)}=\frac{t^3-216}{t^2(t^2-36)}$$
   $$\frac{3t}{t^2-36}=\frac{3t^3}{t^2(t^2-36)}$$

5. д) $$x^2-9=(x-3)(x+3), \qquad x^2-6x+9=(x-3)^2.$$

   НОЗ: $$ (x-3)^2(x+3).$$

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

6. е) $$a^2+10a+25=(a+5)^2, \qquad a^2-25=(a-5)(a+5).$$

   НОЗ: $$ (a+5)^2(a-5).$$

   $$\frac{a-5}{a^2+10a+25}=\frac{(a-5)^2}{(a+5)^2(a-5)}$$
   $$\frac{a-1}{a^2-25}=\frac{(a-1)(a+5)}{(a-5)(a+5)^2}=\frac{a^2+4a-5}{(a+5)^2(a-5)}$$

Ответ

1. а) $$\frac{(2p-q)^2}{2(2p-q)(2p+q)}, \qquad \frac{2pq}{2(2p-q)(2p+q)}$$

2. б) $$\frac{6x^2}{2(x^2-9y^2)}, \qquad \frac{(x+3y)^2}{2(x^2-9y^2)}$$

3. в) $$\frac{5y^2}{y(y^2-100)}, \qquad \frac{y^3+1000}{y(y^2-100)}$$

4. г) $$\frac{t^3-216}{t^2(t^2-36)}, \qquad \frac{3t^3}{t^2(t^2-36)}$$

5. д) $$\frac{x^2-x-6}{(x-3)^2(x+3)}, \qquad \frac{(x+3)^2}{(x-3)^2(x+3)}$$

6. е) $$\frac{(a-5)^2}{(a+5)^2(a-5)}, \qquad \frac{a^2+4a-5}{(a+5)^2(a-5)}$$

1. а) НОЗ: $$n^2-m^2.$$

   $$\frac{7mn}{n^2-m^2}, \qquad \frac{n^2(n-m)}{n^2-m^2}, \qquad \frac{m^2(n+m)}{n^2-m^2}$$

2. б) НОЗ: $$a^2-9.$$

   $$\frac{a-2}{a^2-9}, \qquad \frac{a^2(a-3)}{a^2-9}, \qquad \frac{3(a+3)}{a^2-9}$$

3. в) $$25-b^2=(5-b)(5+b), \qquad \text{НОЗ: } b^2-25.$$

   $$\frac{-6b}{b^2-25}, \qquad \frac{(b-5)^2}{b^2-25}, \qquad \frac{(b+5)^2}{b^2-25}$$

4. г) НОЗ: $$a^2-x^2.$$

   $$\frac{(a+x)^2}{a^2-x^2}, \qquad \frac{2ax}{a^2-x^2}, \qquad \frac{(a-x)^2}{a^2-x^2}$$

5. д) НОЗ: $$ (a-4)^2(a+4)^2.$$

   $$\frac{(a+4)^2}{(a-4)^2(a+4)^2}, \qquad \frac{a^4-256}{(a-4)^2(a+4)^2}, \qquad \frac{(a-4)^2}{(a-4)^2(a+4)^2}$$

6. е) $$1-y=-(y-1), \qquad 1+y=y+1.$$

   НОЗ: $$ (y-1)^2(y+1)^2.$$

   $$\frac{(y+1)^3}{(y-1)^2(y+1)^2}, \qquad \frac{-(1-y)^3}{(y-1)^2(y+1)^2}, \qquad \frac{(y^2+1)(y^2-1)}{(y-1)^2(y+1)^2}=\frac{y^4-1}{(y-1)^2(y+1)^2}$$