ГДЗ по алгебре 8 класс Макарычев Задание 497

\[\boxed{\text{497\ (497).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\mathbf{Вспомним.}\]

\[Формулы\ квадрата\ разности\ и\ \]

\[квадрата\ суммы:\]

\[\mathbf{a}^{\mathbf{2}}\mathbf{- 2}\mathbf{ab +}\mathbf{b}^{\mathbf{2}}\mathbf{=}\left( \mathbf{a - b} \right)^{\mathbf{2}}\mathbf{;}\]

\[\mathbf{a}^{\mathbf{2}}\mathbf{+ 2}\mathbf{ab +}\mathbf{b}^{\mathbf{2}}\mathbf{=}\left( \mathbf{a + b} \right)^{\mathbf{2}}\mathbf{.}\]

Решение.

\[\textbf{а)}\ при\ x = 1 + \sqrt{5}:\]

\[x^{2} - 6 = \left( 1 + \sqrt{5} \right)^{2} - 6 =\]

\[= 1 + 2\sqrt{5} + 5 - 6 = 2\sqrt{5}.\]

\[\textbf{б)}\ при\ x = 3 - \sqrt{3}:\]

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

\[= \left( 3 - \sqrt{3} \right)\left( 3 - \sqrt{3} - 6 \right) =\]

\[= \left( 3 - \sqrt{3} \right)\left( - \sqrt{3} - 3 \right) =\]

\[= - 3\sqrt{3} - 9 + \sqrt{3}\sqrt{3} + 3\sqrt{3} =\]

\[= - 9 + 3 = - 6.\]

\[\textbf{в)}\ при\ x = 2 + \sqrt{3}:\ \]

\[x^{2} - 4x + 3 =\]

\[= \left( 2 + \sqrt{3} \right)^{2} - 4\left( 2 + \sqrt{3} \right) + 3 =\]

\[= 4 + 4\sqrt{3} + 3 - 8 - 4\sqrt{3} + 3 =\]

\[= 2\]

\[\textbf{г)}\ при\ x = \frac{3 + \sqrt{2}}{2}:\]

\[x^{2} - 3x + 5 =\]

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

\[= \frac{\left( 3 + \sqrt{2} \right)^{2}}{4} - \frac{3\left( 3 + \sqrt{2} \right)}{4} + 5^{\backslash 4} =\]

\[= \frac{9 + 6\sqrt{2} + 2 - 6\left( 3 + \sqrt{2} \right) + 20}{4} =\]

\[= \frac{9 + 6\sqrt{2} + 2 - 18 - 6\sqrt{2} + 20}{4} =\]

\[= \frac{13}{4} = 3\frac{1}{4} = 3,25.\]