Question

1. Укажіть число, що є розв'язком нерівності 3/5 х2 − х − 5/2> 0. ОТВЕТЫ: 0 -1 1 2

Answer 1

$\displaystyle \tt \frac{3}{5}x^2-x-\frac{5}{2}>0 \: \: \: \: \: | \cdot 10\\\displaystyle \tt 6x^2-10x-25>0\\\displaystyle \tt 6x^2-10x-25=0\\\displaystyle \tt x=\frac{5б5\sqrt{7}}{6}$

$\displaystyle \tt 6(x-\frac{5+5\sqrt{7}}{6})(x-\frac{5-\sqrt{7}}{6})>0 \: \: \: \: | \div6\\\displaystyle \tt (x-\frac{5+5\sqrt{7}}{6})(x-\frac{5-\sqrt{7}}{6})>0\\1)\left \{ {{\displaystyle \tt x-\frac{5+5\sqrt{7}}{6}>0} \atop {\displaystyle \tt x-\frac{5-5\sqrt{7}}{6}>0}} \right. \\2)\left \{ {{\displaystyle \tt x-\frac{5+5\sqrt{7}}{6}<0} \atop {\displaystyle \tt x-\frac{5-5\sqrt{7}}{6}<0}} \right.$

$1) \left \{ {\displaystyle \tt {x>\frac{5+5\sqrt{7}}{6}} \atop {\displaystyle \tt x>\frac{5-5\sqrt{7}}{6}}} \right. \\2) \left \{ {{\displaystyle \tt x<\frac{5+5\sqrt{7}}{6}} \atop {\displaystyle \tt x<\frac{5-5\sqrt{7}}{6}}} \right. \\\left \{ {{{\displaystyle \tt x\in (\frac{5+5\sqrt{7}}{6};\: +\infty)} \atop {{\displaystyle \tt x\in (-\infty;\:\frac{5-5\sqrt{7}}{6})}} \right. \\\\\displaystyle \tt x\in (-\infty;\: \frac{5-5\sqrt{7}}{6})\cup(\frac{5+5\sqrt{7}}{6};\: +\infty)$