Question

Chứng minh rằng nếu S = a + b + c thì:
\(S\left(S-2b\right)\left(S-2c\right)+S\left(S-2c\right)\left(S-2a\right)+S\left(S-2a\right)\left(S-2b\right)=\left(S-2a\right)\left(S-2b\right)\left(S-2c\right)+8abc\)

Answer 1

Trả lời giúp bạn nè:
VT = S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b)
= S((S - 2b)(S -2c) + (S-2c)(S - 2a) + (S - 2a)(S - 2b) )
= S ( S2 -2cS -2bS + 4bc + S2 - 2aS - 2cS +4ac + S2 -2bS -2aS +4ab )
= S ( 3S2 - 4cS -4bS - 4aS + 4bc + 4ac + 4ab)
= 3S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S3 + S3 + S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S2 (S -4c ) + S2 (S -4b ) + S2 (S -4a )
= S2 ( S -4c + S - 4b + S - 4a)
= S2 (3S - 4(c + b + a)
= S2 (3S - 4S)
= 3S3 - 4S3
= -S3 ( 1 )

VP = (S - 2a)(S - 2b)(S - 2c) + 8abc
= (S2 -2bS -2aS + 4ab)(S - 2c) + 8abc
= S3 - 2cS2 - 2bS2 + 4bcS - 2aS2 + 4acS + 4abS - 8abc + 8abc
= S3 - 2cS2 - 2bS2 - 2aS2 + 4bcS + 4acS + 4abS
= S2 (S -2c ) - S2 (2b + 2a )
= S2 ( S - 2c - 2b - 2a )
= S2 ( S - 2( c + b + a))
= S3 - 2S3
= -S3 ( 2 )
Từ (1) và (2) suy ra :
S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b) = (S - 2a)(S - 2b)(S - 2c) + 8abc