`39.`
`a)(x-1)³-x(x-3)²+1`
`=x³-3x²+3x-1-x(x²-6x+9)+1`
`=x³-3x²+3x-1-x³+6x²-9x+1`
`=(x³-x³)+(-3x²+6x²)+(3x-9x)+(-1+1)`
`=3x²-6x`
`b)(x+2)³-x²(x+6)`
`=x³+6x²+12x+8-x³-6x²`
`=(x³-x³)+(6x²-6x²)+12x+8`
`=12x+8`
`40.`
`(x+2)³-(x-2)³`
`=x³+6x²+12x+8-(x³-6x²+12x-8)`
`=x³+6x²+12x+8-x³+6x²-12x+8`
`=(x³-x³)+(6x²+6x²)+(12x-12x)+(8+8)`
`=12x²+16`
Vậy biểu thức `(x+2)³-(x-2)³` có phụ thuộc vào biến `x`
`41.`
`a)(x+1)³-x²(x+3)=2`
`⇔x³+3x²+3x+1-x³-3x²=2`
`⇔(x³-x³)+(3x²-3x²)+3x+1=2`
`⇔3x+1=2`
`⇔3x=2-1`
`⇔3x=1`
`⇔x=1/3`
Vậy `x=1/3`
`b)(x-2)³-x(x+1)(x-1)+6x²=5`
`⇔x³-6x²+12x-8-x(x²-1)+6x²=5`
`⇔x³-6x²+12x-8-x³+x+6x²=5`
`⇔(x³-x³)+(-6x²+6x²)+(12x+x)-8=5`
`⇔13x-8=5`
`⇔13x=5+8`
`⇔13x=13`
`⇔x=13:13`
`⇔x=1`
Vậy `x=1`