Наибольшее целое решение неравенства (10-х)*(х^2-14х+40) / х^3-100х >=0

$\frac{(10-x)(x^2-14x+40)}{x^3-100x} \geq 0 \\ \\ \frac{(10-x)(x^2-10x-4x+40)}{x(x^2-100)} \geq 0 \\ \\ \frac{(10-x)(x(x-10)-4(x-10))}{x(x-10)(x+10)} \geq 0 \\ \\ \frac{(10-x)(x-10)(x-4)}{x(x-10)(x+10)} \geq 0 \\ \\ \frac{(10-x)(x-4)}{x(x+10)} \geq 0$
$\frac{-(x-10)(x-4)}{x(x+10)} \geq 0 \\ \\ \frac{(x-10)(x-4)}{x(x+10)} \leq 0$

ОДЗ: x≠0 x≠ -10

x(x-10)(x-4)(x+10)≤0
x=0 x=10 x=4 x= -10
+ - + - +
---------- -10 ----------- 0 ------------ 4 ------------ 10 -------------
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x∈(-10; 0]U[4; 10]
x=10 - наибольшее целое решение неравенства
Ответ: 10.