với các số a,b,c là các số thực thỏa mãn (3a+3b+3c)3 = 24 + (3a+b-c)3 + (3b+c-a)+ (3c+a-b)3

CMR (a+2b) (b+2c) (c+2a) = 1

Đặt $3a+b-c=x, 3b+c-a=y,3c+a-b=z$
$=> 3a+3b+3c=x+y+z$
giả thiết $<=> (x+y+z)^3=24+x^3+y^3+z^3$
$<=>(x+y+z)^3=24+(x+y)^3-3xy(x+y)+z^3$
$<=> (x+y+z)^3=24+(x+y+z)^3-3(x+y)(y+z)(z+x)$
$<=> (x+y)(y+z)(x+z)=8, đpcm$
thanks bạn. mình làm đc rồi –  thanktra 15-06-15 11:46 AM
sau khi đặt như kia thì đpcm ở đề bài tương đương (x y)(y z)(z x)=8 –  phuongthao202 15-06-15 09:43 AM
bạn ơi mình vẫn ko hiểu ra đpcm ntn ? –  thanktra 15-06-15 08:01 AM

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