ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 1287

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\[Зададим\ І\ число\ p + t\sqrt{2},\]

\[\ \ а\ ІІ\ число\ x + y\sqrt{2}:\ \]

\[Сумма:\ \ \ p + t\sqrt{2} + x + y\sqrt{2} =\]

\[= (p + x) + \sqrt{2} \cdot (t + y) =\]

\[= a + b\sqrt{2}\]

\[a = x + p,\ \ b = y + t.\]

\[Разность:\ p + t\sqrt{2} - x - y\sqrt{2} =\]

\[= (p - x) + \sqrt{2} \cdot (t - y) =\]

\[= a + b\sqrt{2}\]

\[a = p - x,\ \ b = t - y.\]

\[Произведение:\ \]

\[\left( p + t\sqrt{2} \right)\left( x + y\sqrt{2} \right) = px +\]

\[+ py\sqrt{2} + tx\sqrt{2} + 2ty =\]

\[= (px + 2ty) + \sqrt{2} \cdot (py + tx) =\]

\[= a + b\sqrt{2}\]

\[a = px + 2ty,\ \ b = py + xt.\]

\[Частное:\ \]

\[\frac{p + t\sqrt{2}}{x + y\sqrt{2}} =\]

\[= \frac{\left( p + t\sqrt{2} \right)\left( x - y\sqrt{2} \right)}{\left( x + y\sqrt{2} \right)\left( x - y\sqrt{2} \right)} =\]

\[= \frac{\left( p + t\sqrt{2} \right)\left( x - y\sqrt{2} \right)}{x^{2} - 2y^{2}} =\]

\[= \frac{xp - yp\sqrt{2} + tx\sqrt{2} - yt \cdot 2}{x^{2} - 2y^{2}} =\]

\[= \frac{(xp - 2ty) - \sqrt{2} \cdot (yp - tx)}{x^{2} - 2y^{2}} =\]

\[= \frac{xp - 2ty}{x^{2} - 2y^{2}} - \frac{\sqrt{2} \cdot (yp - tx)}{x^{2} - 2y^{2}} =\]

\[= a + b\sqrt{2}\]

\[a = \frac{xp - 2ty}{x^{2} - 2y^{2}},\]

\[\ \ b = \frac{(yp - tx)}{x^{2} - 2y^{2}}\]