Răspuns :

a) [tex](4x-2)^2-(4x+3)(4x-3)<1-10x <=> 16x^2-16x+4- \\ -(16x^2-9) < 1-10x <=>16x^2-16x+4-16x^2+9<1- \\ -10x<=>-16x+13<1-10x <=>16x-10x>13-1 <=> \\ <=> 6x>12 <=> x>2[/tex]

b) [tex] \frac{(x-5)^2-9}{x^2+5x-14} = \frac{(x-5-3)(x-5+3)}{x^2-2x+7x-14}= \frac{(x-8)(x-2)}{x(x-2)+7(x-2)}= \frac{(x-8)(x-2)}{(x+7)(x-2)}= \frac{x-8}{x+7} [/tex]