Question

Решите пожалуйста /( ^_^) /​

Answer 1

$\left \{ {{((x+5)(x-5)-(x+3)x>x+1} \atop {(3x+2)^2-(4-3x)^2\leq 14+37x}} \right. \\\left \{ {{x^2-25-x^2-3x-x>1} \atop {9x^2+12x+4-16+24x-9x^2\leq 14+37x}} \right. \\\left \{ {{-4x>1+25} \atop {36x-37x\leq 14+12}} \right. \left \{ {{-4x>26,|:(-4)} \atop {-x\leq26|\cdot(-1) }} \right. \\\left \{ {{x<-6,5} \atop {x\geq -26}} \right. \\\\x\in [-26; -6,5).$

Наибольшее целое решение - х = -6.