ОДЗ:
x²-10x+9>0
x²-10x+9=0
D=100-36=64
x₁=(10-8)/2=1
x₂=(10+8)/2=9
+ - +
--------- 1 ------------- 9 ------------
\\\\\\\\\\\ \\\\\\\\\\\\\
x∈(-∞; 1)U(9; +∞)
Так как 1/2<1, то
x² -10x+9≤(1/2)⁰
x² -10x+9≤1
x² -10x+9-1≤0
x² -10x+8≤0
x² -10+8=0
D=100-32=68
x₁=(10-√68)/2=5-√17≈ 0.88
x₂=5+√17≈ 9.12
+ - +
--------- 5-√17 -------------- 5+√17 -----------
\\\\\\\\\\\\\\\\
x∈[5-√17; 5+√17]
Объединяем два множества:
-------- 5-√17 ---- 1 ---------------- 9 -------5+√17 ------------
\\\\\\\\\\ \\\\\\\\\\
х∈[5-√17; 1)U(9; 5+√17]
Ответ: [5-√17; 1)U(9; 5+√17]