Bài `13`:
`M_{(x)}=2x^4+3x^2-x+1-x^2-x^4-6x^3`
`M_{(x)}=(2x^4-x^4)-6x^3+(3x^2-x^2)-x+1`
`M_{(x)}=x^4-6x^3+2x^2-x+1`
Bài `14`:
Ta có:
$\widehat{C}=180^0-\widehat{A}-\widehat{B}$
$\widehat{C}=180^0-80^0-60^0$
$\widehat{C}=40^0$
Vì `80^0 > 60^0 > 40^0` nên `BC > AC > AB`
