Решебник по алгебре 9 класс Макарычев Задание 362

 

\[\boxed{\text{362\ (362).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x^{4} - 6x^{2} + 3 = 0\]

\[x = \sqrt{3 + \sqrt{5}}:\]

\[\left( \sqrt{3 + \sqrt{5}} \right)^{4} - 6 \cdot\]

\[\cdot \left( \sqrt{3 + \sqrt{5}} \right)^{2} + 3 = 0\]

\[\left( 3 + \sqrt{5} \right)^{2} - 6 \cdot \left( 3 + \sqrt{5} \right) + 3 = 0\]

\[9 + 6\sqrt{5} + 5 - 18 - 6\sqrt{5} + 3 = 0\]

\[- 1 \neq 0 \Longrightarrow \sqrt{3 + \sqrt{5}} - не\ \]

\[является\ корнем.\]

\[\textbf{б)}\ x^{4} - 10x^{2} + 23 = 0\]

\[x = \sqrt{5 - \sqrt{2}}:\]

\[\left( \sqrt{5 - \sqrt{2}} \right)^{4} - 10 \cdot\]

\[\cdot \left( \sqrt{5 - \sqrt{2}} \right)^{2} + 23 = 0\]

\[\left( 5 - \sqrt{2} \right)^{2} - 10 \cdot \left( 5 - \sqrt{2} \right) +\]

\[+ 23 = 0\]

\[25 - 10\sqrt{2} + 2 - 50 + 10\sqrt{2} +\]

\[+ 23 = 0\]

\[0 = 0 \Longrightarrow \sqrt{5 - \sqrt{2}} - является\ \]

\[корнем.\]

\[\boxed{\text{362.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ y = kx + b;\ \ A(0;1);B( - 2;0)\]

\[1 = 0k + b\]

\[b = 1.\]

\[0 = - 2k + 1\]

\[2k = 1\]

\[k = 0,5.\]

\[y = 0,5x + 1\]

\[y - 0,5x = 1.\]

\[\textbf{б)}\ y = kx + b;\ \ A( - 2;0);\ \ \]

\[B(0; - 1)\]

\[- 1 = 0k + b\]

\[b = - 1.\]

\[0 = - 2k - 1\]

\[- 2k = 1\]

\[k = - 0,5.\]

\[y = - 0,5x - 1\]

\[y + 0,5x = - 1.\]

\[\textbf{в)}\ y = - 1.\]