Question

x^2-3x+4>0 и x^2-16<0​

Answer 1

${x}^{2} - 3x + 4 > 0 \\ x = r$

${x}^{2} - 16 < 0 \\ ( x + 4)(x - 4) < 0 \\ x > - 4 \\ x < 4$