Đáp án:
2.xa+2b+cxa+2b+c =y2a+b−cy2a+b−c =z4a−4b+cz4a−4b+c =t
Ta có: t=x+2y+z(a+2b+c)+2(2a+b−c)+(4a−4b+c)x+2y+z(a+2b+c)+2(2a+b−c)+(4a−4b+c) =x+2y+z9ax+2y+z9a (1)
t=2x+y−z2(a+2b+c)+(2a+b−c)−(4a−4b+c)2x+y−z2(a+2b+c)+(2a+b−c)−(4a−4b+c) =2x+y−z9b2x+y−z9b (2)
t=4x−4y+z4(a+2b+c)−4(2a+b−c)+(4a−4b+c)4x−4y+z4(a+2b+c)−4(2a+b−c)+(4a−4b+c) =4x−4y+z9c4x−4y+z9c (3)
Từ (1),(2),(3) ta có: x+2y+z9ax+2y+z9a=2x+y−z9b2x+y−z9b =4x−4y+z9c4x−4y+z9c =t
⇔ (Đpcm)
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