Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1335

\[\boxed{\mathbf{1335}\mathbf{.}}\]

\[1)\ x^{2} + ax - b^{2} + \frac{a^{2}}{4} = 0\]

\[x^{2} + ax - \left( b^{2} - \frac{a^{2}}{4} \right) = 0\]

\[D = a^{2} + 4\left( b^{2} - \frac{a^{2}}{4} \right) =\]

\[= a^{2} + 4b^{2} - a^{2} = 4b^{2}\]

\[x = \frac{- a \pm \sqrt{4b^{2}}}{2} = \frac{- a \pm 2b}{2} =\]

\[= - \frac{a}{2} \pm b.\]

\[Ответ:\ \ x = - \frac{a}{2} \pm b.\]

\[2x(2x + a) - x(2x - a) = 5a^{2}\]

\[4x^{2} + 2ax - 2x^{2} + ax - 5a^{2} = 0\]

\[2x^{2} + 3ax - 5a^{2} = 0\]

\[D = (3a)^{2} + 4 \bullet 2 \bullet 5a^{2} =\]

\[= 9a^{2} + 40a^{2} = 49a^{2}\]

\[x_{1} = \frac{- 3a - 7a}{2 \bullet 2} = - \frac{10a}{4} = - 2,5a;\]

\[x_{2} = \frac{- 3a + 7a}{2 \bullet 2} = \frac{4a}{4} = a.\]

\[Ответ:\ \ x_{1} = - 2,5a;\ \ x_{2} = a.\]