Упр.912 ГДЗ Колягин Ткачёва 8 класс (Алгебра)

1) (x^6+x^4-x^2-1)/(x^3+x^2+x+1);
2) (x^3+x^2-4x-4)/(x^3-3x-2);
3) (x^4-2x^3+x-2)/(x^3-3x^2+3x-2);
4) (x^3+5x^2+7x+3)/(2x^3+5x^2+4x+1);
5) (x^4-16)/(x^4-4x^3+8x^2-16x+16).

1. $$\frac{x^6+x^4-x^2-1}{x^3+x^2+x+1}= \frac{x^4(x^2+1)-(x^2+1)}{x^2(x+1)+(x+1)}= \frac{(x^2+1)(x^4-1)}{(x+1)(x^2+1)}$$

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

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

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

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

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

4. $$\frac{x^3+5x^2+7x+3}{2x^3+5x^2+4x+1}= \frac{x^3+5x^2+2x+5x+2+1}{x^3+x^3+4x^2+x^2+4x+1}$$

   $$=\frac{(x^3+1)+5x(x+1)+2(x+1)}{(x^3+1)+x^2(x+1)+4x(x+1)}$$

   $$=\frac{(x+1)(x^2-x+1)+5x(x+1)+2(x+1)}{(x+1)(x^2-x+1)+x^2(x+1)+4x(x+1)}$$

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

5. $$\frac{x^4-16}{x^4-4x^3+8x^2-16x+16}= \frac{(x^2-4)(x^2+4)}{x^2(x^2+4)-4x(x^2+4)+4(x^2+4)}$$

   $$=\frac{(x-2)(x+2)(x^2+4)}{(x^2+4)(x^2-4x+4)}= \frac{(x-2)(x+2)}{(x-2)^2}=\frac{x+2}{x-2}.$$

Ответ

1) $$ (x-1)(x^2+1) $$;
2) $$ \frac{x+2}{x+1} $$;
3) $$ x+1 $$;
4) $$ \frac{x+3}{2x+1} $$;
5) $$ \frac{x+2}{x-2} $$.