\[А\] | \[Б\] | \[В\] |
---|---|---|
\[3\] | \[2\] | \[4\] |
\[- 7;\ - 5;\ - 3;\ldots\]
\[a_{1} = - 7;\ \ a_{2} = - 5:\]
\[d = - 5 - ( - 7) = 2.\]
\[S_{6} = \frac{2a_{1} + d(n - 1)}{2} \cdot n =\]
\[= \frac{2 \cdot ( - 7) + 2 \cdot (6 - 1)}{2} \cdot 6 =\]
\[= ( - 14 + 10) \cdot 3 = - 4 \cdot 3 = - 12.\]
\[Ответ:\ - 12.\]
\[\frac{16x^{2}}{4x - 1} + \frac{1}{1 - 4x} - 4x - 4 =\]
\[= \frac{16x^{2}}{4x - 1} - \frac{1}{4x - 1} - (4x + 4)^{\backslash 4x - 1} =\]
\[= \frac{16x^{2} - 1 - 16x^{2} - 16x + 4x + 4}{4x - 1} =\]
\[= \frac{- 12x + 3}{4x - 1} = \frac{- 3 \cdot (4x - 1)}{4x - 1} = - 3.\]
\[Ответ:\ - 3.\]
\[\frac{2x - 3}{x + 1} \geq 1^{\backslash x + 1}\]
\[\frac{2x - 3 - x - 1}{x + 1} \geq 0;\ \ \ x \neq - 1\]
\[\frac{x - 4}{x + 1} \geq 0\]
\[x < - 1;\ \ \ x \geq 4.\]
\[Ответ:4.\]
\[Ответ:3.\]