Question

9. Rezolvă:
a) (x + 1)(x + 5) < x2 + 17, unde a E R;
b) (x + 1)(x + 5) +17, unde € N
; c) 22 - 20 > (+4)2, unde E N;
d) 2 - 20 > (2 + 4), unde eR;
e) (x - 2)2 – (x – 3)(x + 1)
Ajutor vă rog​

Answer 1

Răspuns:

A) x²+5x+x+5<2x+17

X²+6x+5<2x+17

X²+6x-2x<17-5

X²+4X<12 /:4

X²+x<3

B) x²+5x+x+5+17

X²+6x+22

C) 22-20>8 /+20

22>30 /:2.

11>15

D) 2-20>2+4

2-20>6 /+20

2>26 /:2

13

E) 2x-4-(x²+x-3x-3)

2x-4-x²+2x+3

X²+4x-1

Explicație pas cu pas:

Sper ca te am ajutat!!