\({10^{\left( {x + 1} \right)\left( {3x + 4} \right)}} + 9 \cdot {10^{\left( {x + 1} \right)\left( {x + 2} \right)}} = {10^{1-x-{x^2}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{10^{3{x^2} + 7x + 4}} + 9 \cdot {10^{{x^2} + 3x + 2}} = {10^{1-x-{x^2}}}\left| { \cdot {{10}^{x + {x^2}}} \ne 0\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\,\,{10^{4{x^2} + 8x + 4}} + 9 \cdot {10^{2{x^2} + 4x + 2}} = 10.\)
Пусть \({10^{2{x^2} + 4x + 2}} = t\), где \(t > 0\). Тогда:
\({t^2} + 9t-10 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -10 < 0,}\\{t = 1.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)
Возвращаясь к прежней переменной, получим:
\({10^{2{x^2} + 4x + 2}} = 1\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,{10^{2{x^2} + 4x + 2}} = {10^0}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\,2{x^2} + 4x + 2 = 0\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,{x^2} + 2x + 1 = 0\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,x = -1.\)
Ответ: \(-1\).