a: Đặt P(x)=0
=>2x+3=0
hay x=-3/2
b: Q(x)=0
=>1-x3=0
hay x=1
`P(x) = 2x + 3`
`2x+ 3 = 0`
`2x = -3`
`x = -3 : 2`
`x = (-3)/2`
_________________
`Q(x) = 1- x^3`
` 1 - x^3 = 0`
` x^3 = 1`
a) cho P(x) = 0
`=> 2x+3 =0`
`=> 2x =-3`
`=> x=-3/2`
Vậy `x=-3/2`
cho Q(x) = 0
`=>1-x^3 =0`
`=> x^3 =1`
`=>x = 1`
vậy `x=1`