toán 8

H

huy14112

$4(a^2)+4ab+4(b^2)-12a-12b+12=(2a+b-3)^3+3(b-1)^2 \ge 0$

Dấu = xảy ra $\leftrightarrow$

$b-1=0 \rightarrow b=1$

$ 2a+b-3=0 $

$2a-2=0$

$a=b=1$


 
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