a)
`A(x) = `$x^{3}$ `+` $3x^{2}$ `+ 7 - 5x`
`= `$x^{3}$ `+` $3x^{2}$ `- 5x + 7`
`B(x) = `$x^{3}$ `+` $3x^{2}$ `- 3 + 12x`
`= `$x^{3}$ `+` $3x^{2}$ `+ 12x -3`
b)
`A(x)+B(x) = `$x^{3}$ `+` $3x^{2}$ `- 5x + 7` `+` $x^{3}$ `+` $3x^{2}$ `+ 12x -3`
`= (`$x^{3}$ `+`$x^{3}$`) + (`$3x^{2}$`+`$3x^{2}$`)+(-5x+12x)+(7-3)`
`=` $2x^{3}$ `+` $6x^{2}$ `+ 7x + 4`
`A(x)-B(x) = `$x^{3}$ `+` $3x^{2}$ `- 5x + 7` `-(` $x^{3}$ `+` $3x^{2}$ `+ 12x -3)`
`= `$x^{3}$ `+` $3x^{2}$ `- 5x + 7` `-` $x^{3}$ `-` $3x^{2}$ `- 12x +3`
`= (`$x^{3}$ `-`$x^{3}$`) + (`$3x^{2}$`-`$3x^{2}$`)+(-5x-12x)+(7+3)`
`= -17x + 10`
c)
`C(x)=A(x)-B(x)=-17x + 10`
Ta có: `C(x)=0`
`=> -17x + 10 = 0`
`=> -17x = -10`
`=> x=10/17`