Решебник по алгебре 7 класс Макарычев ФГОС Задание 540

 

\[\boxed{\text{540.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

Пояснение.

Чтобы узнать, какие числа являются корнями, подставим их в уравнения и вычислим. Если равенство верное, то число – корень уравнения.

Решение.

\[\textbf{а)}\ x^{4} = 81\]

\[x = - 3:\]

\[( - 3)^{4} = 81\]

\[81 = 81.\]

\[x = - 2:\]

\[( - 2)^{4} = 81\]

\[16 \neq 81.\]

\[x = - 1:\]

\[( - 1)^{4} = 81\]

\[1 \neq 81.\]

\[x = 1:\]

\[1^{4} = 81\]

\[1 \neq 81.\]

\[x = 2:\]

\[2^{4} = 81\]

\[16 \neq 81.\]

\[x = 3:\]

\[3^{4} = 81\]

\[81 = 81.\]

\[Ответ:x = \pm 3.\]

\[\textbf{б)}\ x^{6} = 64\]

\[x = - 3:\]

\[( - 3)^{6} = 64\]

\[729 \neq 64.\]

\[x = - 2:\]

\[( - 2)^{6} = 64\]

\[64 = 64.\]

\[x = - 1:\]

\[( - 1)^{6} = 64\]

\[1 \neq 64.\]

\[x = 1:\]

\[1^{6} = 64\]

\[1 \neq 64.\]

\[x = 2:\]

\[2^{6} = 64\]

\[64 = 64.\]

\[x = 3:\]

\[3^{6} = 64\]

\[729 \neq 64.\]

\[Ответ:x = \pm 2.\]

\[\textbf{в)}\ x^{2} - x = 2\]

\[x = - 3:\]

\[( - 3)^{2} - ( - 3) = 2\]

\[9 + 3 = 2\]

\[11 \neq 2.\]

\[x = - 2:\]

\[( - 2)^{2} - ( - 2) = 2\]

\[4 + 2 = 2\]

\[6 \neq 2.\]

\[x = - 1:\]

\[( - 1)^{2} - ( - 1) = 2\]

\[1 + 1 = 2\]

\[2 = 2.\]

\[x = 1:\]

\[1^{2} - 1 = 2\]

\[1 - 1 = 2\]

\[0 \neq 2.\]

\[x = 2:\]

\[2^{2} - 2 = 2\]

\[4 - 2 = 2\]

\[2 = 2.\]

\[x = 3:\]

\[3^{2} - 3 = 2\]

\[9 - 3 = 2\]

\[6 \neq 2.\]

\[x = 2;\ \ \ x = - 1\]

\[Ответ:x = 2;\ \ x = - 1.\]

\[\textbf{г)}\ x^{4} + x^{3} = 6x^{2}\ \ \ |\ :x^{2}\]

\[x^{2} + x = 6\]

\[x = - 3:\]

\[( - 3)^{2} - 3 = 6\]

\[9 - 3 = 6\]

\[6 = 6.\]

\[x = - 2:\]

\[( - 2)^{2} - 2 = 6\]

\[4 - 2 = 6\]

\[2 \neq 6.\]

\[x = - 1:\]

\[( - 1)^{2} - 1 = 6\]

\[1 - 1 = 6\]

\[0 \neq 6.\]

\[x = 1:\]

\[1^{2} + 1 = 6\]

\[1 + 1 = 6\]

\[2 \neq 6.\]

\[x = 2:\]

\[2^{2} + 2 = 6\]

\[4 + 2 = 6\]

\[6 = 6.\]

\[x = 3:\]

\[3^{2} + 3 = 6\]

\[9 + 3 = 6\]

\[12 \neq 6.\]

\[x = - 3;\ \ x = 2\]

\[Ответ:x = - 3;x = 2.\]

\[\textbf{д)}\ x^{3} - 3x^{2} - 4x + 12 = 0\]

\[x = - 3:\]

\[( - 3)^{3} - 3 \cdot ( - 3)^{2} - 4 \cdot ( - 3) + 12 =\]

\[= 0\]

\[- 27 - 27 + 12 + 12 = 0\]

\[- 54 + 24 = 0\]

\[- 30 \neq 0.\]

\[x = - 2:\]

\[( - 2)^{3} - 3 \cdot ( - 2)^{2} - 4 \cdot ( - 2) + 12 =\]

\[= 0\]

\[- 8 - 12 + 8 + 12 = 0\]

\[0 = 0.\]

\[x = - 1:\]

\[( - 1)^{3} - 3 \cdot ( - 1)^{2} - 4 \cdot ( - 1) + 12 =\]

\[= 0\]

\[- 1 - 3 + 4 + 12 = 0\]

\[12 \neq 0.\]

\[x = 1:\]

\[1^{3} - 3 \cdot 1^{2} - 4 \cdot 1 + 12 = 0\]

\[1 - 3 - 4 + 12 = 0\]

\[6 \neq 0.\]

\[x = 2:\]

\[2^{3} - 3 \cdot 2^{2} - 4 \cdot 2 + 12 = 0\]

\[8 - 12 - 8 + 12 = 0\]

\[0 = 0.\]

\[x = 3:\]

\[3^{3} - 3 \cdot 3^{2} - 4 \cdot 3 + 12 = 0\]

\[27 - 27 - 12 + 12 = 0\]

\[0 = 0.\]

\[Ответ:x = \pm 2;\ \ x = 3.\]

\[\textbf{е)}\ x^{3} + 3x^{2} - x - 3 = 0\]

\[x = - 3:\]

\[( - 3)^{3} + 3 \cdot ( - 3)^{2} - ( - 3) - 3 =\]

\[= 0\]

\[- 27 + 27 + 3 - 3 = 0\]

\[0 = 0.\]

\[x = - 2:\]

\[( - 2)^{3} + 3 \cdot ( - 2)^{2} - ( - 2) - 3 =\]

\[= 0\]

\[- 8 + 12 + 2 - 3 = 0\]

\[3 \neq 0.\]

\[x = - 1:\]

\[( - 1)^{3} + 3 \cdot ( - 1)^{2} - ( - 1) - 3 =\]

\[= 0\]

\[- 1 + 3 + 1 - 3 = 0\]

\[0 = 0.\]

\[x = 1:\]

\[1^{3} + 3 \cdot 1^{2} - 1 - 3 = 0\]

\[1 + 3 - 1 - 3 = 0\]

\[0 = 0.\]

\[x = 2:\]

\[2^{3} + 3 \cdot 2^{2} - 2 - 3 = 0\]

\[8 + 12 - 2 - 3 = 0\]

\[15 \neq 0.\]

\[x = 3:\]

\[3^{3} + 3 \cdot 3^{2} - 3 - 3 = 0\]

\[27 + 27 - 3 - 3 = 0\]

\[48 \neq 0.\]

\[x = 1;\ \ x = - 1;\ \ x = - 3.\]

\[Ответ:x = \pm 1;\ \ x = - 3.\ \]

\[\boxed{\text{540\ (540).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Если\ радиус\ увеличить\ \]

\[в\ 2\ раза,\ то:\ \]

\[V = \frac{4}{3}\pi \cdot (2r)^{3} = \frac{4}{3}\pi \cdot 8r^{3} =\]

\[= 8 \cdot \left( \frac{4}{3}\pi r^{3} \right)\]

\[то\ есть,\ объем\ увеличится\ \]

\[в\ 8\ раз.\]

\[Если\ радиус\ увеличить\ \]

\[в\ 4\ раза,\ то:\ \]

\[V = \frac{4}{3}\pi \cdot (4r)^{3} = \frac{4}{3}\pi \cdot 64r^{3} =\]

\[= 64 \cdot \left( \frac{4}{3}\pi r^{3} \right)\ \]

\[то\ есть,\ объем\ увеличится\ \]

\[в\ 64\ раза.\]