Решебник по алгебре 9 класс Мерзляк Задание 754

 

\[\boxed{\mathbf{754\ (754).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ a,\ b,\ c;\ \ \ \ b = \frac{a + c}{2};\ \ \]

\[2b = a + c\]

\[a^{2} + 8bc = (2b + c)^{2}\]

\[a^{2} + 4 \cdot (a + c) \cdot c =\]

\[= (a + c + c)^{2}\]

\[a^{2} + 4ac + 4c^{2} = (a + 2c)^{2}\]

\[a^{2} + 4ac + 4c^{2} =\]

\[= a^{2} + 4ac + 4c^{2}\]

\[0 = 0 \Longrightarrow доказано.\]

\[2)\ \frac{2}{9} \cdot (a + b + c)^{3} = a^{2}(b + c) +\]

\[+ b^{2}(a + c) + c²(a + b)\]

\[\frac{2}{9} \cdot (2b + b)^{3} = a^{2}\left( \frac{a + c}{2} + c \right) +\]

\[+ b^{2} \cdot 2b + c^{2}\left( a + \frac{a + c}{2} \right)\]

\[\frac{2}{9} \cdot (3b)^{3} = a^{2}\left( \frac{a + 3c}{2} \right) + 2b^{3} +\]

\[+ c^{2}\left( \frac{3a + c}{2} \right)\]

\[\frac{2}{9} \cdot 27b^{3} = a^{2}\left( \frac{a + 3c}{2} \right) +\]

\[+ 2b^{3} + c^{2}\left( \frac{3a + c}{2} \right)\]

\[6b^{3} =\]

\[= \frac{a^{2}(a + 3c) + c^{2}(3a + c)}{2} +\]

\[+ 2b^{3}\]

\[6b^{3} = \frac{a^{3} + 3a^{2}c + 3ac^{2} + c^{3}}{2} +\]

\[+ 2b^{3}\]

\[6b^{3} = \frac{(a + c)^{3}}{2} + 2b^{3}\]

\[6b^{3} = \frac{(2b)^{3}}{2} + 2b^{3}\]

\[6b^{3} = 4b^{3} + 2b^{3}\]

\[6b³ = 6b³ \Longrightarrow доказано.\]

\[\boxed{\mathbf{754.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[y =\]

\[= \underset{\begin{matrix} бесконечная\ геометрическая \\ прогрессия \\ \end{matrix}}{\overset{\sqrt{x} + \frac{\sqrt{x}}{1 + \sqrt{x}} + \frac{\sqrt{x}}{\left( 1 + \sqrt{x} \right)^{2}} + \ldots}{︸}};\]

\[\ \ x > 0\]

\[q = \frac{\sqrt{x}}{1 + \sqrt{x}}\ :\sqrt{x} = \frac{1}{1 + \sqrt{x}},\]

\[S = \frac{\sqrt{x}}{1 - \frac{1}{1 + \sqrt{x}}} = \frac{\sqrt{x}}{\frac{1 + \sqrt{x} - 1}{1 + \sqrt{x}}} =\]

\[= \frac{\sqrt{x} \cdot (1 + \sqrt{x})}{\sqrt{x}} = 1 + \sqrt{x}\]

\[y = 1 + \sqrt{x};\ \ x > 0;\ \ x \neq 0\]

\[график\ y = \sqrt{x}\ \ перенести\ на\ 1\]

\[\ единицу\ вверх.\]