Задача 14. Решите уравнение \(\sqrt {1-{2^{x + 1}} + {4^x}} = {2^{2x + 3}}-10 \cdot {2^x} + 2\)
ОТВЕТ: -3; 0.
\(\sqrt {1-{2^{x + 1}} + {4^x}} = {2^{2x + 3}}-10 \cdot {2^x} + 2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sqrt {1-2 \cdot {2^x} + {4^x}} = 8 \cdot {2^{2x}}-10 \cdot {2^x} + 2.\) Пусть \({2^x} = t\), где \(t > 0\). Тогда: \(\sqrt {1-2t + {t^2}} = 8{t^2}-10t + 2\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\sqrt {{{\left( {t-1} \right)}^2}} = 8{t^2}-10t + 2\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left| {t-1} \right| = 8{t^2}-10t + 2\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{t-1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{t-1 = 8{t^2}-10t + 2}\end{array}\,\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{t-1 < 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{-t + 1 = 8{t^2}-10t + 2}\end{array}} \right.}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{t \ge 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{8{t^2}-11t + 3 = 0}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{t < 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{8{t^2}-9t + 1 = 0}\end{array}} \right.}\end{array}} \right.} \right.\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{t \ge 1,}\\{\left[ {\begin{array}{*{20}{c}}{t = 1,}\\{t = \dfrac{3}{8}}\end{array}} \right.}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{t < 1,}\\{\left[ {\begin{array}{*{20}{c}}{t = 1,}\\{t = \dfrac{1}{8}}\end{array}} \right.}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,}\\{t = \dfrac{1}{8}.}\end{array}} \right.\) Возвращаясь к прежней переменной, получим: \(\left[ {\begin{array}{*{20}{c}}{{2^x} = 1,}\\{{2^x} = \dfrac{1}{8}}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{2^x} = {2^0},}\\{{2^x} = {2^{-3}}}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 0,\,\,\,}\\{x = -3.}\end{array}} \right.} \right.\) Ответ: \(-3;\,\,\,\,0\).