Страница 204 номер 577, ГДЗ по алгебре за 9 класс к учебнику Алимова с ответами и решением
Сократить дробь: 1) \(\frac{4 - b^2}{4b + 2b^2}\); 2) \(\frac{b^2 - 9}{3b^2 - 9b}\); 3) \(\frac{5a^2 - 10ab}{ab - 2b^2}\); 4) \(\frac{3xy - 21y^2}{4x^2 - 28xy}\); 5) \(\frac{x^2 - x - 12}{x^2 - 16}\); 6) \(\frac{x^2 - x - 20}{x^2 - 25}\); 7) \(\frac{3x^2 - 2x - 8}{2x^2 - 3x - 2}\); 8) \(\frac{2x^2 + x - 3}{2x^2 + 7x + 6}\).
1) \(\frac{4 - b^{2}}{4b + 2b^{2}} = \frac{(2 - b)(2 + b)}{2b(2 + b)} = \frac{2 - b}{2b}\);
2) \(\frac{b^{2} - 9}{3b^{2} - 9b} = \frac{(b - 3)(b + 3)}{3b(b - 3)} = \frac{b + 3}{3b}\);
3) \(\frac{5a^{2} - 10ab}{ab - 2b^{2}} = \frac{5a(a - 2b)}{b(a - 2b)} = \frac{5a}{b}\);
4) \(\frac{3xy - 21y^{2}}{4x^{2} - 28xy} = \frac{3y(x - 7y)}{4x(x - 7y)} = \frac{3y}{4x}\);
5) \(\frac{x^{2} - x - 12}{x^{2} - 16} = \frac{(x - 4)(x + 3)}{(x - 4)(x + 4)} = \frac{x + 3}{x + 4}\);
\(x_{1{,}2} = \dfrac{1 \pm \sqrt{1 + 48}}{2},\)
\(x_{1} = 4, x_{2} = -3;\)
6) \(\frac{x^{2} - x - 20}{x^{2} - 25}\)
\(x_{1{,}2} = \dfrac{1 \pm \sqrt{1 + 80}}{2},\)
\(x_{1} = 5, x_{2} = -4;\)
\(\dfrac{x^{2} - x - 20}{x^{2} - 25} = \dfrac{(x - 5)(x + 4)}{(x - 5)(x + 5)} = \dfrac{x + 4}{x + 5};\)
7) \(\frac{3x^{2} - 2x - 8}{2x^{2} - 3x - 2}\)
\(x_{1{,}2} = \dfrac{2 \pm \sqrt{4 + 96}}{6},\)
\(x_{1} = 2, x_{2} = -\dfrac{4}{3};\)
\(x_{3{,}4} = \dfrac{3 \pm \sqrt{9 + 16}}{4},\)
\(x_{3} = 2, x_{4} = -\dfrac{1}{2};\)
\(\dfrac{3x^{2} - 2x - 8}{2x^{2} - 3x - 2} = \dfrac{3(x - 2)\left(x + \frac{4}{3}\right)}{2(x - 2)\left(x + \dfrac{1}{2}\right)} = \dfrac{3x + 4}{2x + 1};\)
8) \(\frac{2x^{2} + x - 3}{2x^{2} + 7x + 6}\)
\(x_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 24}}{4},\)
\(x_{1} = 1, x_{2} = -\dfrac{3}{2};\)
\(x_{3{,}4} = \dfrac{-7 \pm \sqrt{49 - 48}}{4},\)
\(x_{3} = -\dfrac{3}{2}, x_{4} = -2;\)
\(\dfrac{2x^{2} + x - 3}{2x^{2} + 7x + 6} = \dfrac{(x - 1)\left(x + \frac{3}{2}\right)}{(x + 2)\left(x + \dfrac{3}{2}\right)} = \dfrac{x - 1}{x + 2}.\)
