Đáp án:
Giải thích các bước giải:
`a,(y+3)^3-(y+1)^3=56`
`<=>[(y+3)-(y+1)][(y+3)^2+(y+3)(y+1)+(y+1)^2]-56=0`
`<=>2(y^2+6y+9+y^2+4y+3+y^2+2y+1)-56=0`
`<=>y^2+6y+9+y^2+4y+3+y^2+2y+1-28=0`
`<=>3y^2+12y-15=0`
`<=>y^2+4y-5=0`
`<=>(y-1)(y+5)=0`
`<=>`\(\left[ \begin{array}{l}y=1\\y=-5\end{array} \right.\)
`b,(y-2)^3-(y-3)(y^2+3y+9)+6(y+1)^2=49`
`<=>y^3-6y^2+12y-8-y^3+27+6y^2+12y+6-49=0`
`<=>24y-24=0`
`<=>y-1=0`
`<=>y=1`
`c,(x-1)^3+1+3x(x-4)=0`
`<=>(x-1+1)[(x-1)^2-(x-1)+1]+3x(x-4)=0`
`<=>x(x^2-2x+1-x+1+1]+3x(x-4)=0`
`<=>x(x^2-2x+1-x+1+1+3x-12)=0`
`<=>x(x^2-9)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x^2-9=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\\left[ \begin{array}{l}x=3\\x=-3\end{array} \right.\end{array} \right.\)
`d,12x-x^2-36=0`
`<=>x^2-12x+36=0`
`<=>(x-6)^2=0`
`<=>x-6=0`
`<=>x=6`
