giúp mình nha :))) huhuhuuu

Đáp án:

Giải thích các bước giải:

 a) `(y-2)^3-(y-3)(y^2+3y+9)+6(y+1)^2=49`

`⇔ y^3-6y^2+12y-8-(y^3-27)+6(y^2+2y+1)-49=0`

`⇔ y^3-6y^2+12y-8-y^3+27+6y^2+12y+6-49=0`

`⇔ 24y-24=0`

`⇔ y=24:24`

`⇔ y=1`

Vậy `S={1}`

b) `(y+3)^3-(y+1)^3=56`

`⇔ y^3+9y^2+27y+27-(y^3+3y^2+3y+1)-56=0`

`⇔ y^3+9y^2+27y+27-y^3-3y^2-3y-1-56=0`

`⇔ 6y^2+24y-30=0`

`⇔ 6y^2-6y+30y-30=0`

`⇔ 6y(y-1)+30(y-1)=0`

`⇔ (6y+30)(y-1)=0`

`⇔` \(\left[ \begin{array}{l}6y+30=0\\y-1=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}y=-5\\y=1\end{array} \right.\) 

Vậy `S={1;-5}`