[tex]\log_{0,5}(x^2+0,5x)\geq1, \\ \left \{ {{x^2+0,5x>0,} \atop {x^2+0,5x\leq1;}} \right. \ \left \{ {{x(x+0,5)>0,} \atop {x^2+0,5x-1\leq0;}} \right. \\ x(x+0,5)=0, x_1=0, x_2=-0,5; \\ x\in(-\infty;-0,5)\cup(0;+\infty); \\[/tex]
[tex]x^2+0,5x-1=0, \\ D=4,25, x_1=\frac{-1-\sqrt{17}}{4}\approx-1,3, x_2=\frac{-1+\sqrt{17}}{4}\approx0,8, \\ x^2+0,5x-1=(x-\frac{-1-\sqrt{17}}{4})(x-\frac{-1-\sqrt{17}}{4}), \\ (x-\frac{-1-\sqrt{17}}{4})(x-\frac{-1-\sqrt{17}}{4})\leq0, \\ \frac{-1-\sqrt{17}}{4}\leq x\leq\frac{-1+\sqrt{17}}{4}, \\ x\in(\frac{-1-\sqrt{17}}{4};-0,5)\cup(0;\frac{-1+\sqrt{17}}{4}) [/tex]
