CM a^2 - 3ab + 3b^2 > hoặc = 0

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.`