Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 265

\[\boxed{\mathbf{265}\mathbf{.}}\]

\[1)\ 3^{|x - 2|} < 9\]

\[3^{|x - 2|} < 3^{2}\]

\[|x - 2| < 2\]

\[\sqrt{(x - 2)^{2}} < 2\]

\[(x - 2)^{2} < 4\]

\[x^{2} - 4x + 4 < 4\]

\[x^{2} - 4x < 0\]

\[x(x - 4) < 0\]

\[0 < x < 4\]

\[Ответ:\ \ 0 < x < 4.\]

\[2)\ 4^{|x + 1|} > 16\ \]

\[4^{|x + 1|} > 4^{2}\ \]

\[|x + 1| > 2\ \]

\[\sqrt{(x + 1)^{2}} > 2\ \]

\[(x + 1)^{2} > 4\ \]

\[x^{2} + 2x + 1 > 4\ \]

\[x^{2} + 2x - 3 > 0\ \]

\[D = 2^{2} + 4 \bullet 3 = 4 + 12 = 16\]

\[x_{1} = \frac{- 2 - 4}{2} = - 3;\text{\ \ }\]

\[x_{2} = \frac{- 2 + 4}{2} = 1.\ \]

\[(x + 3)(x - 1) > 0\ \]

\[x < - 3;\text{\ \ }x > 1\ \]

\[Ответ:\ \ x < - 3;\ \ \ x > 1.\]

\[3)\ 2^{|x - 2|} > 4^{|x + 1|}\ \]

\[2^{|x - 2|} > 2^{2|x + 1|}\ \]

\[|x - 2| > 2|x + 1|\ \]

\[\sqrt{(x - 2)^{2}} > 2\sqrt{(x + 1)^{2}}\ \]

\[(x - 2)^{2} > 4(x + 1)^{2}\ \]

\[x^{2} - 4x + 4 > 4\left( x^{2} + 2x + 1 \right)\ \]

\[x^{2} - 4x + 4 > 4x^{2} + 8x + 4\ \]

\[3x^{2} + 12x < 0\ \]

\[x^{2} + 4x < 0\ \]

\[(x + 4)x < 0\ \]

\[- 4 < x < 0\ \]

\[Ответ:\ \ - 4 < x < 0.\]

\[4)\ 5^{|x + 4|} < 25^{|x|}\ \]

\[5^{|x + 4|} < 5^{2|x|}\ \]

\[|x + 4| < 2|x|\ \]

\[\sqrt{(x + 4)^{2}} < 2\sqrt{x^{2}}\ \]

\[(x + 4)^{2} < 4x^{2}\ \]

\[x^{2} + 8x + 16 < 4x^{2}\ \]

\[3x^{2} - 8x - 16 > 0\ \]

\[D = 8^{2} + 4 \bullet 3 \bullet 16 =\]

\[= 64 + 192 = 256\]

\[x_{1} = \frac{8 - 16}{2 \bullet 3} = - \frac{8}{6} = - \frac{4}{3} = - 1\frac{1}{3}\ \]

\[x_{2} = \frac{8 + 16}{2 \bullet 3} = \frac{24}{6} = 4\ \]

\[\left( x + 1\frac{1}{3} \right)(x - 4) > 0\ \]

\[x < - 1\frac{1}{3};\ \ x > 4\ \]

\[Ответ:\ \ x < - 1\frac{1}{3};\ \ \ x > 4.\]