ГДЗ по алгебре 7 класс Макарычев Задание 392

\[\boxed{\text{392\ (392).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ если\ x = - 9:\]

\[\ x^{2} = ( - 9)^{2} = 81;\ \ \]

\[- x^{2} = - ( - 9)^{2} = - 81;\ \ \]

\[( - x)^{2} = 9^{2} = 81.\]

\[если\ x = 9:\]

\[\ x^{2} = (9)^{2} = 81;\ \ \]

\[- x^{2} = - 9^{2} = - 81;\ \ \]

\[( - x)^{2} = ( - 9)^{2} = 81.\]

\[если\ x = - 6:\]

\[\ x^{2} = ( - 6)^{2} = 36;\ \ \]

\[- x^{2} = - ( - 6)^{2} = - 36;\ \]

\[\ ( - x)^{2} = 6^{2} = 36.\]

\[если\ x = 6:\]

\[\ x^{2} = (6)^{2} = 36;\ \ \]

\[- x^{2} = - 6^{2} = - 36;\ \ \]

\[( - x)^{2} = {( - 6)}^{2} = 36.\]

\[если\ x = - 2:\]

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

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

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

\[если\ x = 2:\]

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

\[- x^{2} = - 2^{2} = - 4;\ \ \]

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

\[\textbf{б)}\ если\ x = - 4:\]

\[x^{3} = ( - 4)^{3} = - 64;\ \ \ \ \]

\[- x^{3} = - ( - 4)^{3} = 64;\ \ \]

\[\ ( - x)^{3} = 4^{3} = 64.\]

\[если\ x = 4:\]

\[x^{3} = (4)^{3} = 64;\ \ \ \ \]

\[- x^{3} = - 4^{3} = - 64;\ \ \ \]

\[( - x)^{3} = ( - {4)}^{3} = - 64.\]

\[если\ x = - 3:\]

\[x^{3} = ( - 3)^{3} = - 27;\ \ \ \]

\[\ - x^{3} = - ( - 3)^{3} = 27;\ \ \ \]

\[( - x)^{3} = 3^{3} = 27.\]

\[если\ x = 3:\]

\[x^{3} = (3)^{3} = 27;\ \ \ \ \]

\[- x^{3} = - 3^{3} = - 27;\ \ \]

\[\ ( - x)^{3} = ( - 3)^{3} = - 27.\]

\[если\ x = - 1:\]

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

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

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

\[если\ x = 1:\]

\[x^{3} = (1)^{3} = 1;\ \ \ \]

\[\ - x^{3} = - 1^{3} = - 1;\ \ \]

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