Question

Решите неравенство 2√7(12-5х)+3√3(5х-12)≥0.

Answer 1

$2 \sqrt{7}(12-5x)+3 \sqrt{3}(5x-12)=24 \sqrt{7} -10 \sqrt{7} *x+15 \sqrt{3}*x-36 \sqrt{3}=$

$x(15 \sqrt{3}-10 \sqrt{7}) \geq 36 \sqrt{3}-24 \sqrt{7}$

$x \geq \frac{12(3 \sqrt{3}-2 \sqrt{7} ) }{5(3 \sqrt{3}-2 \sqrt{7}) }$

$x \geq \frac{12}{5}$

$x \geq 2,4$

Ответ [2,4 +беск[