Решебник по алгебре 9 класс Макарычев Задание 751

 

\[\boxed{\text{751\ (751).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{a - 3}{a^{2} - 3a + 9} - \frac{6a - 18}{a^{3} + 27} =\]

\[= \frac{(a - 3)(a + 3) - 6 \cdot (a - 3)}{a^{3} + 27} =\]

\[= \frac{(a - 3)(a + 3 - 6)}{a^{3} + 27} =\]

\[= \frac{(a - 3)(a - 3)}{a^{3} + 27}\]

\[2)\ \frac{(a - 3)(a - 3)}{a^{3} + 27} \cdot \frac{4 \cdot \left( a^{3} + 27 \right)}{5 \cdot (a - 3)} =\]

\[= \frac{4}{5} \cdot (a - 3).\]

\[\textbf{б)}\ \frac{ab^{2} - a^{2}b}{a + b} \cdot \frac{a + \frac{\text{ab}}{a - b}}{a - \frac{\text{ab}}{a + b}} =\]

\[= \frac{\text{ab}(b - a)}{a + b} \cdot \frac{1 + \frac{b}{a + b}}{1 - \frac{b}{a + b}} =\]

\[= \frac{\text{ab}(b - a)}{a + b} \cdot \frac{\frac{a - b + b}{a - b}}{\frac{a + b - b}{a + b}} =\]

\[= \frac{\text{ab}(b - a)}{a + b} \cdot \frac{a}{a - b} \cdot \frac{a + b}{a} =\]

\[= - ab.\]

\[\boxed{\text{751.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\frac{1}{8}x³ = 1\]

\[x^{3} = 8\]

\[x³ = 2³\]

\[x = 2\]

\[Ответ:x = 2.\]

\[\textbf{б)}\ 1000x³ + 1 = 0\]

\[x^{3} = - \frac{1}{1000}\]

\[x^{3} = ( - 0,1)^{3}\]

\[x = - 0,1\]

\[Ответ:\ x = - 0,1.\]

\[\textbf{в)}\frac{1}{27}x³ = 0,001\]

\[\left( \frac{x}{3} \right)^{3} = {0,1}^{3}\]

\[\frac{x}{3} = 0,1\]

\[x = 0,3\]

\[Ответ:x = 0,3.\]

\[\textbf{г)}\frac{1}{9}x^{4} - 16 = 0\]

\[x^{4} = 144\]

\[x^{2} = 12\]

\[x = \pm \sqrt{12} = \pm 2\sqrt{3}\]

\[Ответ:\ x = \pm 2\sqrt{3}.\]

\[\textbf{д)}\ 1 + x^{5} = 0\]

\[x^{5} = ( - 1)^{5}\]

\[x = - 1\]

\[Ответ:\ x = - 1.\]

\[\textbf{е)}\ x^{8} - 16 = 0\]

\[x^{8} = 16\]

\[\left( x^{2} \right)^{4} = 2^{4}\]

\[x^{2} = 2\]

\[x = \pm \sqrt{2}\]

\[Ответ:\ x = \pm \sqrt{2}.\]