Question

7-2x / x^2-12x+32>=1/4

Answer 1

x²-12x+32=0

D/4=36-32=2²

x1=6-2=4; x2=6+2=8 ОДЗ х≠4;  х≠8

(7-2х)/(х-4)(х-8)≥1/4   I *4

(7-2х)*4/(х-4)(х-8)  -  1 ≥ 0

[4(7-2х)-(х-4)(х-8)]/(x-4)(x-8)≥0

[28-8x-x²+12x-32]/(x-4)(x-8)≥0

[-x²+4x-4]/(x-4)(x-8)≥0

(x-2)²/(4-x)(x-8)≥0

----------[2]-------(4)-----(8)----->x

   -             -          +         -

x∈(4; 8).