Question

корень из (24-10x)>3-4x
Прошу помочь, никак не могу понять как решать

Answer 1

$\sqrt{24-10x}>3-4x\; \; \Leftrightarrow \; \; \; \left \{ {{3-4x\geq 0\qquad \qquad } \atop {24-10x>9-24x+16x^2}} \right.\; \; ili\; \; \left \{ {{3-4x<0\; \; } \atop {24-10x\geq 0}} \right.\\\\\left \{ {{4x\leq 3\qquad \qquad } \atop {16x^2-14x-15<0}} \right.\; \; \; ili\; \; \; \left \{ {{4x>3} \atop {10x\leq 24}} \right.\\\\\left \{ {{x\leq \frac{3}{4}\qquad \quad } \atop {(x+\frac{5}{8})(x-\frac{3}{2})<0}} \right.\; \; \; \; ili\; \; \; \; \left \{ {{x>\frac{3}{4}} \atop {x\leq \frac{12}{5}}} \right.$

$\left \{ {{x\leq \frac{3}{4}\qquad } \atop {x\in (-\frac{5}{8}\, ,\, \frac{3}{2})}} \right. \; \; \; \; \; ili\; \; \; \; \; x\in (\, \frac{3}{4}\, ,\, \frac{12}{5}\, ]\\\\x\in (-\frac{5}{8}\, ,\, \frac{3}{4}\, ]\cup (\frac{3}{4}\, ,\, \frac{12}{5}\, ]=(-\frac{5}{8}\, ,\, \frac{12}{5}\, ]\\\\Otvet:\; \; x\in (-0,625\, ;\, 2,4\, ]\; .$