ГДЗ по алгебре 9 класс Мерзляк Задание 1028

\[\boxed{\mathbf{1028\ (1028).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ a,\ b,\ c;\ \ b = a + d,\]

\[\ \ c = a + 2d,\]

\[\text{\ \ }то\ \ \ ac = a \cdot (a + 2d).\]

\[R = \frac{\text{abc}}{4S} - радиус\ описанной\ \]

\[окружности;\]

\[r = \frac{S}{p} - радиус\ вписанной\]

\[\ окружности.\]

\[6Rr = 6 \cdot \frac{\text{abc}}{4S} \cdot \frac{S}{p} = \frac{6}{4} \cdot \frac{\text{abc}}{p} =\]

\[= \frac{3}{2} \cdot \frac{\text{abc}}{p} =\]

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

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

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

\[= \frac{3a(a + d)(a + 2d)}{3 \cdot (a + d)} =\]

\[= a(a + 2d) \Longrightarrow \ ac = 6Rr.\]

\[2)\ b = a + d;\ \ c = a + 2d,\ \]

\[По\ т.\ Пифагора\ a² + b² = c²\]

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

\[a^{2} + a^{2} + 2ad + d^{2} =\]

\[= a^{2} + 4ad + 4d^{2}\]

\[3d^{2} + 2ad - a^{2} = 0\]

\[D = 4a^{2} + 12a^{2} = 16a^{2}\]

\[d = \frac{- 2a + 4a}{6} = \frac{a}{3}\]

\[d = \frac{- 2a - 4a}{6} =\]

\[= - a\ (не\ удовлетворяет)\]

\[d = \frac{a}{3},\ \ тогда\ \ b = a + d =\]

\[= a + \frac{a}{3} = \frac{4}{3}a\]

\[c = a + 2d = a + \frac{2a}{3} = \frac{5a}{3}\]

\[r = \frac{a + b - c}{2} = \frac{a + \frac{4}{3}a - \frac{5}{3}a}{2} =\]

\[= \frac{\frac{2}{3}a}{2} = \frac{a}{3} \Longrightarrow \ r = d = \frac{a}{3}.\]

\[3)\ По\ т.\ косинусов:(a + 2d)^{2} =\]

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

\[- - 2a(a + d) \cdot cos120{^\circ}.\]

\[a^{2} + 4ad + 4d^{2} = a^{2} + a^{2} +\]

\[+ 2ad + d^{2} - 2a(a + d) \cdot \left( - \frac{1}{2} \right)\]

\[a^{2} + 4ad + 4d^{2} = 2a^{2} +\]

\[+ 2ad + d^{2} + a^{2} + ad\]

\[3a^{2} + ad - 2a^{2} = 0\]

\[D = a^{2} + 24a^{2} = 25a^{2}\]

\[d = \frac{- a + 5a}{6} = \frac{2}{3}a\]

\[d = \frac{- a - 5a}{6} =\]

\[= - a\ (не\ удовлетворяет).\]

\[d = \frac{2}{3}a,\ \ то\ \ \ a + d = a +\]

\[+ \frac{2}{3}a = \frac{5}{3}a,\]

\[\ \ a + 2d = a + \frac{4}{3}a = \frac{7}{3}a\]

\[Тогда\ \ a\ :\frac{5}{3}\text{a\ }:\frac{7}{3}a = 3\ :5\ :7.\]