Question

решите неравенство: (1-3x)^2<=5(1-3x)

Answer 1

$(1-3x)^2le5(1-3x),\1-6x+9x^2le5-15x,\9x^2+9x-4le0,\\9x^2+9x-4=0,\D=9^2-4bullet9bullet(-4)=81+144=225,\\x_1=frac{-9+sqrt{225}}{18}=frac{1}{3}, x_2=frac{-9-sqrt{225}}{18}=frac{-24}{18}=frac{-4}{3}=-1frac{1}{3},\\9(x-frac{1}{3})(x+1frac{1}{3})le0,\\ -1frac{1}{3}le xlefrac{1}{3}.$

Ответ: $xin[-1frac{1}{3}; frac{1}{3}].$