Question

Решите неравенство: (1-3x)^2< =5(1-3x)

Answer 1

$(1-3x)^2\le5(1-3x),\\1-6x+9x^2\le5-15x,\\9x^2+9x-4\le0,\\\\9x^2+9x-4=0,\\D=9^2-4\bullet9\bullet(-4)=81+144=225,\\\\x_1=\frac{-9+\sqrt{225}}{18}=\frac{1}{3},\ x_2=\frac{-9-\sqrt{225}}{18}=\frac{-24}{18}=\frac{-4}{3}=-1\frac{1}{3},\\\\9(x-\frac{1}{3})(x+1\frac{1}{3})\le0,\\\\ -1\frac{1}{3}\le x\le\frac{1}{3}.$

ответ: $x\in[-1\frac{1}{3};\ \frac{1}{3}].$