`a) => 3(x-5)(x^2-4)=7+3x^3-15x^2`
`=> 3(x^3-4x-5x^2+20)=7+3x^3-15x^2`
`=> 3x^3-12x-15x^2+60=7+3x^3-15x^2`
`=> 3x^3-12x-15x^2+60-7-3x^3+15x^2=0`
`=> -12x+53=0`
`=> -12x=-53`
`=> x=53/12`
`b) => 16(2-3x)-x^2(2-3x)=0`
`=> (2-3x)(16-x^2)=0`
`=> (2-3x)(4-x)(4+x)=0`
`=> 2-3x=0` hoặc `4-x=0` hoặc `4+x=0`
`=> x=2/3` hoặc `x=4` hoặc `x=-4`
`c) => x^3-7x^2-7+x=0`
`=> x^2(x-7)+(x-7)=0`
`=> (x-7)(x^2+1)=0`
`=> x-7=0` hoặc `x^2+1=0`
`=> x=7` hoặc `x^2=-1 (loại)`
`Vậy` `x=7`