Đáp án:

 

Giải thích các bước giải:

 TA có:

`a)6x^2-72x=0`

`6x.(x-12)=0`

`x.(x-12)=0`

`=>[(x=0),(x-12=0):}`

`=>[(x=0),(x=0+12):}`

`=>[(x=0),(x=12):}`

Vậy `x in {0;12}`

`---`

`b)-2x^4+16x=0`

`-2x.(x^3-8)=0`

`x.(x^3-8)=0`

`=>[(x=0),(x^3-8=0):}`

`=>[(x=0),(x^3=8):}`

`=>[(x=0),(x^3=2^3):}`

`=>[(x=0),(x=2):}`

Vậy `x in {0;2}`

`---`

`c)x(x-5)-(x-3)^2=0`

`x^2-5x-(x^2-6x+9)=0`

`x^2-5x-x^2+6x-9=0`

`x-9=0`

`x=0+9`

`x=9`

Vậy `x=9`

`---`

`d)(x-2)^3-(x-2)(x^2+2x+4)=0`

`(x^3-6x^2+12x-8)-(x-2)(x^2+2.x+2^2)=0`

`x^3-6x^2+12x-8-x^3+8=0`

`-6x^2+12x=0`

`-6x.(x-2)=0`

`x.(x-2)=0`

`=>[(x=0),(x-2=0):}`

`=>[(x=0),(x=0+2):}`

`=>[(x=0),(x=2):}`

Vậy `x in {0;2}`