Решебник по алгебре 8 класс Макарычев ФГОС Задание 300

 

\[\boxed{\text{300\ (300).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \sqrt{81} = 9\]

\[9 > 0;\ \ \ 9^{2} = 81.\]

\[\textbf{б)}\ \sqrt{36} = 6\]

\[6 > 0;\ \ \ 6^{2} = 36.\]

\[\textbf{в)}\ \sqrt{1600} = 40\]

\[40 > 0;\ \ 40^{2} = 1600.\]

\[\textbf{г)}\ \sqrt{10\ 000} = 100\]

\[100 > 0;\ \ \ 100^{2} = 10\ 000.\]

\[\textbf{д)}\ \sqrt{0,04} = 0,2\]

\[0,2 > 0;\ \ (0,2)^{2} = 0,04.\]

\[\textbf{е)}\ \sqrt{0,81} = 0,9\]

\[0,9 > 0;\ \ \ {0,9}^{2} = 0,81.\]

\[\textbf{ж)}\ \sqrt{\frac{81}{4}} = \frac{9}{2}.\]

\[\frac{9}{2} > 0;\ \ \ \left( \frac{9}{2} \right)^{2} = \frac{81}{4}.\]

\[\textbf{з)}\ \sqrt{1\frac{24}{25}} = \sqrt{\frac{49}{25}}\ = \frac{7}{5}.\]

\[\frac{7}{5} > 0;\ \ \ \left( \frac{7}{5} \right)^{2} = \frac{49}{25} = 1\frac{24}{25}.\]

\[\boxed{\text{300.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[\textbf{а)}\ A\ \left( \sqrt{15,21} \right) = A\ (3,9),\ \ \]

\[\ \text{B\ }\left( - \sqrt{16} \right) = B\ ( - 4)\]

\[Ближе\ точка\ \text{A\ }\left( \sqrt{15,21} \right).\]

\[\textbf{б)}\ A\ \left( \sqrt{2\frac{7}{9}} \right) = A\ \left( \sqrt{\frac{25}{9}} \right) =\]

\[= A\ \left( \frac{5}{3} \right);\ \ \text{B\ }\left( - \sqrt{1\frac{13}{36}} \right) =\]

\[= B\ \left( - \sqrt{\frac{49}{36}} \right) = B\ \left( - \frac{7}{6} \right)\ \]

\[Ближе\ точка\ B\ \left( - \sqrt{1\frac{13}{36}} \right).\]