\[t\ (ч)\] |
\[Производ\] \[\frac{(дет\ }{ч})\] |
\[А\ (деталей)\] | |
---|---|---|---|
\[Мастер\] | \[\frac{72}{x + 4}\ на\ 2\ ч < \searrow\] | \[x + 4\] | \[72\] |
\[Ученик\] | \[\frac{64}{x}\] | \[x\] | \[64\] |
\[\mathbf{Составим\ уравнение:}\]
\[\frac{64}{x} - \frac{72}{x + 4} = 2\]
\[\frac{64 \cdot (x + 4) - 72x}{x(x + 4)} = 2\]
\[64x + 256 - 72x = 2x^{2} + 8x\]
\[2x^{2} + 8x + 8x - 256 = 0\]
\[2x^{2} + 16x - 256 = 0\ \ \ \ \ \ \ |\ :2\]
\[x^{2} + 8x - 128 = 0\]
\[D = b^{2} - 4ac = 64 - 4 \cdot 1 \cdot ( - 128) =\]
\[= 64 + 512 = 576\]
\[x_{1} = \frac{- 8 + 24}{2} = \frac{16}{2} = 8\ (деталей) -\]
\[ученик.\]
\[x_{2} = \frac{- 8 - 24}{2} = \frac{- 32}{2} = - 16 < 0 \Longrightarrow \ не\ \]
\[подходит.\]
\[1)\ 8 + 4 = 12\ (деталей) - мастер.\]
\[Ответ:\ \ в\ час\ \ мастер\ изготовил\ 12\]
\(\ деталей\ и\) \(ученик\ изготовил\ 8\ деталей.\)
\[y = x^{2} - 4x + 3\]
\[1)\ x_{0} = \frac{- b}{2a} = \frac{4}{2} = 2\]
\[y_{0}(2) = 4 - 8 + 3 = - 1 \Longrightarrow (2;\ - 1).\]
\[2)\ y = 0 \Longrightarrow\]
\[x^{2} - 4x + 3 = 0\]
\[x_{1} + x_{2} = 4\]
\[x_{1} \cdot x_{2} = 3 \Longrightarrow x_{1} = 3\ \ \ \ и\ \ \]
\[x_{2} = 1 \Longrightarrow (3;0)\ и\ (1;0).\]
\[3)\ x = 0 \Longrightarrow\]
\[y = 0^{2} - 4 \cdot 0 + 3 = 3 \Longrightarrow (0;3).\]
\[6x^{2} - 7x - 24 < 0\]
\[6x² - 7x - 24 = 0\]
\[D = b^{2} - 4ac = 49 - 4 \cdot 6 \cdot ( - 24) =\]
\[= 49 + 576 = 625\]
\[x_{1} = \frac{7 + 25}{12} = \frac{32}{12} = \frac{8}{3} = 2\frac{2}{3}\]
\[x_{2} = \frac{7 - 25}{12} = - \frac{18}{12} = - \frac{3}{2} = - 1,5\]
\[Ответ:x \in \left( - 1,5;2\frac{2}{3} \right)\text{.\ }\]
\[\sqrt{28} \cdot \left( \sqrt{14} - \sqrt{7} \right) - 2\sqrt{98} =\]
\[= \sqrt{28 \cdot 14} - \sqrt{28 \cdot 7} - 2\sqrt{98} =\]
\[= \sqrt{4 \cdot 7 \cdot 7 \cdot 2} - \sqrt{4 \cdot 7 \cdot 7} - 2\sqrt{49 \cdot 2} =\]
\[= 14\sqrt{2} - 14 - 14\sqrt{2} = - 14.\]
\[\left\{ \begin{matrix} x - 4y = 3\ \ \ \ \ \ \\ x^{2} - 21y = 28 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = 4y + 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (4y + 3)^{2} - 21y = 28 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 4y + 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 16y^{2} + 24y + 9 - 21y = 28 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = 4y + 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 16y² + 3y - 19 = 0 \\ \end{matrix} \right.\ \]
\[16y^{2} + 3y - 19 = 0\]
\[D = b^{2} - 4ac = 9 - 4 \cdot 16 \cdot ( - 19) =\]
\[= 9 + 1216 = 1225\]
\[y_{1} = \frac{- 3 + 35}{32} = \frac{32}{32} = 1\]
\[y_{2} = \frac{- 3 - 35}{32} = - \frac{38}{32} = - \frac{19}{16} = - 1\frac{3}{16}\]
\[\left\lbrack \begin{matrix} \left\{ \begin{matrix} y = 1 \\ x = 7 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ } \\ \left\{ \begin{matrix} y = - 1\frac{3}{16} \\ x = - 1,75 \\ \end{matrix} \right.\ \\ \end{matrix} \right.\ \]
\[Ответ:(7;1)\ \ и\ \ \ \left( - 1,75;\ - 1\frac{3}{16} \right)\text{.\ }\]