Пожалуйста помогите решить неравенство

$\frac{x^2-8x+11}{x-6}+ \frac{x^2-9x+2}{x-9} \leq 2x-2 \\ \\ \frac{(x^2-8x+11)(x-9)+(x^2-9x+2)(x-6)-(2x-2)(x-6)(x-9)}{(x-6)(x-9)} \leq 0 \\ \\ \frac{x^3-8x^2+11x-9x^2+72x-99+x^3-9x^2+2x-6x^2+54x-12-(2x^2-2x-12x+12)(x-9)}{(x-6)(x-9)} \leq 0 \\ \\ \frac{x^3-17x^2+83x-99+x^3-15x^2+56x-12-(2x^3-14x^2+12x-18x^2+126x-108)}{(x-6)(x-9)} \leq 0 \\ \\ \frac{x^3+x^3-2x^3-17x^2-15x^2+14x^2+18x^2+83x+56x-12x-126x-99-12+108}{(x-6)(x-9)} \leq 0 \\ \\ \frac{x-3}{(x-6)(x-9)} \leq 0 \\ \\$

x≠6
x≠9

(x-3)(x-6)(x-9)≤0
x=3 x=6 x=9
- + - +
--------- 3 ----------- 6 --------- 9 -----------
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x∈(-∞; 3]U(6; 9)

Ответ: (-∞; 3]U(6; 9).