\({4^{x + 0,5}} + {4^{0,5-x}}-7 \cdot {2^x}-7 \cdot {2^{-x}} + 9 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,2 \cdot \left( {{4^x} + \dfrac{1}{{{4^x}}}} \right)-7 \cdot \left( {{2^x} + \dfrac{1}{{{2^x}}}} \right) + 9 = 0.\)
Пусть \({2^x} + \dfrac{1}{{{2^x}}} = t\). Возведём обе части в квадрат:
\({\left( {{2^x} + \dfrac{1}{{{2^x}}}} \right)^2} = {t^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{4^x} + 2 \cdot {2^x} \cdot \dfrac{1}{{{2^x}}} + \dfrac{1}{{{4^x}}} = {t^2}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{4^x} + \dfrac{1}{{{4^x}}} = {t^2}-2.\)
Тогда уравнение примет вид:
\(2\left( {{t^2}-2} \right)-7t + 9 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2{t^2}-7t + 5 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = 1,\,\,}\\{t = \dfrac{5}{2}.}\end{array}} \right.\)
Вернёмся к прежней неизвестной: \(\left[ {\begin{array}{*{20}{c}}{{2^x} + \dfrac{1}{{{2^x}}} = 1,\,\,}\\{{2^x} + \dfrac{1}{{{2^x}}} = \dfrac{5}{2}.}\end{array}} \right.\)
Пусть \({2^x} = y\), где \(y > 0\). Тогда:
\(\left[ {\begin{array}{*{20}{c}}{y + \dfrac{1}{y} = 1,}\\{y + \dfrac{1}{y} = \dfrac{5}{2}}\end{array}\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{y^2}-y + 1 = 0,\,\,\,\,\,\,}\\{2{y^2}-5y + 2 = 0}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,} \right.\,\,\left[ {\begin{array}{*{20}{c}}{y = \dfrac{1}{2},}\\{y = 2.}\end{array}} \right.\)
Возвращаясь к переменной x, получим: \(\left[ {\begin{array}{*{20}{c}}{{2^x} = \dfrac{1}{2},}\\{{2^x} = 2\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = 1.\,\,\,}\end{array}} \right.} \right.\)
Ответ: \( \pm 1\).