Đề 7:
`1.`
`VT=(a-b)³+3ab(a-b)`
`=a³-3a²b+3ab²-b³+3a²b-3ab²`
`=a³+(-3a²b+3a²b)+(3ab²-3ab²)-b³`
`=a³-b³`
`=VP`
`⇒đpcm`
`2.`
`(x-3)³-(x+3)³`
`=x³-3.x².3+3.x.3²-3³-(x³+3.x².3+3.x.3²+3³)`
`=x³-9x²+27x-27-(x³+9x²+27x+27)`
`=x³-9x²+27x-27-x³-9x²-27x-27`
`=(x³-x³)-(9x²+9x²)+(27x-27x)-(27+27)`
`=-18x²-54`
`3.`
`VT=a³-b³`
`=a³-b³-3a²b+3a²b-3ab²+3ab²`
`=(a³-3a²b+3ab²-b³)+(3a²b-3ab²)`
`=(a-b)³+3ab(a-b)`
`=1³+3ab.1`
`=1+3ab`
`=VP`
`⇒đpcm`
Đề 8:
`1.`
`(1/2a+b)^3+(1/2a-b)^3`
`=(1/2a)^3+3. (1/2a)^2 .b+3. 1/2a .b²+b³+[(1/2a)^3-3. (1/2a)^2 .b+3. 1/2a .b²-b³]`
`=1/8a³+3/4a²b+3/2ab²+b³+1/8a³-3/4a²b+3/2ab²-b³`
`=(1/8a³+1/8a³)+(3/4a²b-3/4a²b)+(3/2ab²+3/2ab²)+(b³-b³)`
`=1/4a³+3ab²`
`2.`
`x³-3x²+3x-1=0`
`⇔x³-3.x².1+3.x.1²-1³=0`
`⇔(x-1)³=0`
`⇔x-1=0`
`⇔x=1`
Vậy `x=1`
`3.`
`(4x-1)³-(4x-3)(16x²+3)`
`=(4x)³-3.(4x)².1+3.4x.1²-1³-[4x(16x²+3)-3(16x²+3)]`
`=64x³-48x²+12x-1-(64x³+12x-48x²-9)`
`=64x³-48x²+12x-1-64x³-12x+48x²+9`
`=(64x³-64x³)+(-48x²+48x²)+(12x-12x)+(-1+9)`
`=8`
Vậy biểu thức trên không phụ thuộc vào `x`
