Question

(x-6)²<(x-2)(x+2) , 3(x+8)>4(8-x) ;

Answer 1

$(x-6)^2 < (x-2)(x+2)\\\\x^2-12x+36 < x^2-4\\\\x^2-12x-x^2 < -4-36\\\\-12x < -40\ \ \ |:12\\\\x > \frac{40}{12}\\\\x > \frac{10}{3}\\\\x > 3\frac{1}{3}\\\\x\in(3\frac{1}3};+\infty)$

$3(x+8) > 4(8-x)\\\\3x+24 > 32-4x\\\\3x+4x > 32-24\\\\7x > 8\ \ \ |:7\\\\x > \frac{8}{7}\\\\x > 1\frac{1}{7}\\\\x\in(1\frac{1}{7};+\infty)$