`a)`
`A=x^2-8x+17`
`=x^2-8x+16+1`
`=x^2-2.x.4+4^2+1`
`=(x-4)^2+1`
Vì `(x-4)^2≥0∀x`
`->(x-4)^2+1≥1∀x`
`->A≥1`
Dấu `'='` xảy ra `<=>x-4=0<=>x=4`
Vậy `A_{min}=1` khi `x=4`
`\text( )`
`B=9x^2+6x+5`
`=9x^2+6x+1+4`
`=(3x)^2+2.3x.1+1^2+4`
`=(3x+1)^2+4`
Vì `(3x+1)^2≥0∀x`
`->(3x+1)^2+4≥4∀x`
`->B≥4`
Dấu `'='` xảy ra `<=>3x+1=0`
`<=>3x=-1<=>x=-1/3`
Vậy `B_{min}=4` khi `x=-1/3`
`b)`
Xét vế trái:
`(a+b+c)^3=[(a+b)+c)^3`
`=(a+b)^3+3.(a+b)^2 .c +3(a+b).c^2+c^3`
`=a^3+3a^2 b +3ab^2+b^3+3.(a+b)^2.c+3(a+b).c^2+c^3`
`=a^3+b^3+c^3+(3a^2 b+3ab^2)+3(a+b)^2 .c+3(a+b).c^2`
`=a^3+b^3+c^3+3ab.(a+b)+3.(a+b).(a+b).c+3(a+b).c^2`
`=a^3+b^3+c^3+3.(a+b).[ab+(a+b).c+c^2]`
`=a^3+b^3+c^3+3(a+b).(ab+ac+bc+c^2)`
`=a^3+b^3+c^3+3(a+b).[a.(b+c)+c(b+c)]`
`=a^3+b^3+c^3+3(a+b).(b+c).(c+a)=VP(đpcm)`
Vậy `(a+b+c)^3=a^3+b^3+c^3+3(a+b).(b+c).(c+a)`
