Question

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Answer 1

ОДЗ
x²-7x+12=0
(x-3)(x-4)=0
x₁=3
x₂=4

3-x=0
x=3

6x-x²≥0
x(6-x)≥0
x(x-6)≤0
x∈[0;6]

$( \frac{1}{(x-3)(x-4)} - \frac{x-4}{x-3}) \sqrt{x(6-x)} \leq 0$
$\frac{1-(x-4)(x-4)}{(x-3)(x-4)} * \sqrt{x(6-x)} \leq 0$
$\frac{1- x^{2}+8x-16 }{(x-3)(x-4)} * \sqrt{x(6-x)} \leq 0$
$\frac{- (x^{2}-8x+15) }{(x-3)(x-4)} * \sqrt{x(6-x)} \leq 0$
$\frac{ (x^{2}-8x+15) }{(x-3)(x-4)} * \sqrt{x(6-x)} \geq 0$
D=8*8-4*15=4
x₁=(8-2)/2=3
x₂=(8+2)/2=5
$\frac{ (x-3)(x-5)) }{(x-3)(x-4)} * \sqrt{x(6-x)} \geq 0$
$\frac{x-5}{x-4}* \sqrt{x(6-x)} \geq 0$
x∈[5; +∞)
учитывая ОДЗ

х∈[5; 6]