2 câu trả lời
$a^2-3ab+3b^2$
$=(a^2-2.a.\dfrac{3}{2}b+\dfrac{9}{4}b^2)+\dfrac{3}{4}b^2$
$=(a-\dfrac{3}{2}b)^2+\dfrac{3}{4}b^2$
Do $(a-\dfrac{3}{2}b)^2>=0,b^2>=0$
$=>(a-\dfrac{3}{2}b)^2+\dfrac{3}{4}b^2>=0$
$=>a^2-3ab+3b^2>=0$
Dấu "=" xảy ra khi: $a=\dfrac{3}{2}b,b=0<=>a=b=0$
Đáp án `+` Giải thích các bước giải:
`a^2 - 3ab + 3b^2`
`=a^2-2 . a . 3/2b+9/4b^2+3/4b^2`
`=(a^2-2 . a . 3/2b+9/4b^2)+3/4b^2`
`=(a-3/2b)^2+3/4b^2`
Do `(a-3/2b)^2>=0 AA a, b`
`->(a-3/2b)^2+3/4b^2>=0 AA a,b`
Hoặc `a^2-3ab+3b^2>=0`(Đpcm)
Dấu "`=`" xảy ra khi:
`(a-3/2b)^2=0, 3/4b^2=0`
`->a-3/2b=0, b^2=0`
`->a=3/2b, b=0`
`->a=b=0.`
