Question

32.18 решите неравенство
2) (4 + y) ^ 2 - y(6 + y) > 0 ;
4) z(z - 10) > (3 - z) ^ 2 .​
дам 15 балов

Answer 1

${(4 + y)}^{2} - y(6 + y) > 0 \\ 16 + 8y + {y}^{2} - 6y - {y}^{2} > 0 \\ 16 + 2y > 0 \\ 2y > 16 \\ y > 8$

$z(z - 10) > (3 - z) {}^{2} \\ {z}^{2} - 10z > 9 - 6z + {z}^{2} \\ - 4z > 9 \\ 4z < - 9 \\ z < - \frac{9}{4} \\ z >-2 \times \frac{1}{4}$