Làm câu c, d thôi ạ

( rút gọn Q = 3/x+1)

`c)`

`M=P.Q`

`=(x^2 +x)/(x+3) . 3/(x+1)`

`=(x(x+1))/(x+3) . 3/(x+1)`

`=(3x)/(x+3)`

`|M| = 3/2`

`=> |(3x)/(x+3)| = 3/2`

`<=> [((3x)/(x+3) = 3/2),((3x)/(x+3) = -3/2):}`

`<=> [(3x.2=3.(x+3)),(3x.2=-3.(x+3)):}`

`<=> [(6x=3x+9),(6x=-3x-9):}`

`<=> [(3x=9),(9x=-9):}`

`<=> [(x=3 (thỏa mãn)),(x=-1 (loại)):}`

Vậy `x=3`.

`d)`

`M=(3x)/(x+3) = (3.(x+3)-9)/(x+3) = 3 - 9/(x+3)`

Vì `x\in NN => x>=0 => x+3 >=3`

`=> 1/(x+3) <= 1/3`

`=> 9/(x+3) <= 9/3 =3`

`=> 3 - 9/(x+3) >= 3-3=0`

`=> M>=0` hay `M_(min) =0`

Dấu "=" xảy ra `<=>x=0`

Vậy `M_(min)=0` khi `x=0`.