ПОМОГИТЕ ПОЖАЛУЙСТА РЕШИТЬ

$(0.5x-2)^{-4}-0.25x^2\ \textgreater \ 4-2x \\Range: x \neq 4\\\frac{1}{(0.5x-2)^{4}}-(0.5x)^2\ \textgreater \ 4-2x \\\frac{1}{(0.5x-2)^{4}}\ \textgreater \ (0.5x)^2-2x+4 \\\frac{1}{(0.5x-2)^{4}}\ \textgreater \ (0.5x-2)^2 \\Substitution: t=(0.5x-2)^2, \ t\ \textgreater \ 0 \\\frac{1}{t^2}\ \textgreater \ t \\\frac{1}{t^2}-t\ \textgreater \ 0 \\\frac{1-t^3}{t^2}\ \textgreater \ 0 \\\frac{(1-t)(1+t+t^2)}{t^2}\ \textgreater \ 0 \\Zeros: \left[\begin{array}{ccc}t=1 \\t^2+t+1=0 \end] \\ \left[\begin{array}{ccc}t=1 \\the\ root\ doesn't\ exist (D=-3) \end]$
$t\ \textless \ 1 \\Reverse\ substitution: (0.5x-2)^2\ \textless \ 1 \\0.25x^2-2x+4-1\ \textless \ 0 \\0.25x^2-2x+3\ \textless \ 0 \\Zeros: 0.25x^2-2x+3=0 \\x^2-8x+12=0 \\\left[\begin{array}{ccc}x=6\\x=2 \end]$
Ответ: x ∈ {(2;4);(4:6)}