Question

Решите пожалуйста неравенство

Answer 1

$sqrt{2x^2-5x-3}>x-1\\1) { left { {{x-1ge0} atop {2x^2-5x-3>(x-1)^2}} right. },\2x^2-5x-3>x^2-2x+1,\2x^2-5x-3-x^2+2x-1>0,\x^2-3x-4>0,\x^2-3x-4=0,\D=9+16=25,\\x_1=frac{3+sqrt{25}}{2}=4, x_2=frac{3-5}{2}=-1,\(x+1)(x-4)>0,\x>4, x<-1,\\ left { {{xge1} atop {x>4, x<-1}} right. , x>4.$

$2) left { {{x-1<0} atop {2x^2-5x-3ge0}} right. ,\\2x^2-5x-3ge0,\2x^2-5x-3=0,\D=25+24=49,\\x_1=frac{5+sqrt{49}}{4}=3, x_2=frac{5-7}{4}=-0.5,\(x+0.5)(x-3)ge0,\xge3, xle-0.5\\ left { {{x<1} atop {xge3, xle-0.5}} right. , xle-0.5$

Ответ: $x>4, xle-0.5$