Question

logx/3(logx(√3-x))≥0

Answer 1

ОДЗ: 0 < x < 3; x ≠ 1

=> x/3 < 1

$log_x \sqrt{3-x} \leq 1\\ 1) 0 \ \textless \ x \ \textless \ 1\\ \sqrt{3 - x} \geq x\\ 3 - x \geq x^2\\ x^2 + x - 3 \leq 0\\ D=1 + 12 = 13\\ x_1 = \frac{ \sqrt{13} -1}{2}\ \textgreater \ \frac{3-1}{2} =1\\ x_2 = \frac{- \sqrt{13}-1 }{2} \ \textless \ 0\\ 0 \ \textless \ x \ \textless \ 1\\\\ 2) 1 \ \textless \ x \ \textless \ 3\\ \sqrt{3 - x} \leq x\\ 3 - x \leq x^2\\ x^2 + x - 3 \geq 0\\ x_1 = \frac{ \sqrt{13} -1}{2}\ \textgreater \ \frac{3-1}{2} =1\\ x_1 \ \textless \ 3\\ x_2 = \frac{- \sqrt{13}-1 }{2} \ \textless \ 0\\ \frac{ \sqrt{13} -1}{2} \leq x \ \textless \ 3$

Ответ: x ∈ (0; 1) U [(√13 - 1)/2; 3)