a) ĐKXD : `x` $\neq$ `-3;±1`
Thay `x=2` vào `P`
`P=(2^2+2)/[3(2+3)]`
`P=6/15`
`P=2/5`
b)
`Q=1/(x-1)+1/(x+1)-(3-x)/(x^2-1)`
⇔`(x+1+x-1-3+x)/(x^2-1)`
⇔`(3x-3)/[(x-1)(x+1)]`
⇔`[3(x-1)]/[(x-1)(x+1)]`
⇔`3/(x+1)`
c)
`P.Q=[3/(x+1)].{(x^2+x)/{3(x+3)]}`
⇔`[3/(x+1)].{[x(x+1)]/[3(x+3)]}`
⇔`x/(x+3)`
`x/(x+3) < 1 ⇔ x/(x+3) - 1 < 0 ⇔ (x-x-3)/(x+3) < 0 ⇔ -3/(x+3) < 0 ⇔ x+3 > 0 ⇔ x> -3`
Vậy `P.Q < 1 ⇔ x> -3` `(x`$\neq$ `-3;±1)`