Namely (x-4) (x+2) >: 0x <; -2 or x> four

g(x)=2x？ + 13x+20≥0

Then (2x+5)(x+4)≥0 x≤-4 or x≥-5/2.

To sum up: x≤-4 or x> four.

(2) f(x)=x？ -2x-8≥(m+2)x-m- 15

x？ -(m+4)x+m+7≥0, established for all x>2.

Let y=x? -(m+4)x+m+7

Y is a parabola with an upward opening, and the vertex is the minimum.

Axis of symmetry x = (m+4)/2 >; 2m & gt； 0

As long as the vertex ≥0 holds, everything else can hold.

So 4(m+7)-(m+4)? ≥0

m？ +4m- 12≤0

-6≤m≤2

All in all: 0

2.∫log4(5)/log3(4)= log4(5)* log4(3)

& lt{[log4 (5)+log4 (3)]/2}？ =( 1/4)[log4 (5*3)]？

& lt 1/4[log4 (4？ )= 1

∴log3 (4) > log4 (5)