Bài 3:
a: \(1+\dfrac{x+4}{5}< x-\dfrac{x+3}{3}\)
=>15+3(x+4)<15x-5(x+3)
=>15+3x+12<15x-5x-15
=>3x+27<10x-15
=>-7x<-42
hay x>6
b: \(\left(x-4\right)\left(x+4\right)>=\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16-x^2-6x-9-5>=0\)
=>-6x-30>=0
=>-6x>30
hay x<=-5