Tính tổng :
(S-2b)(S-2c)+(S-2c)(S-2a)+(S-2a)(S-2b)
Trong đó S= a+b+c
tính tổng T=(s-2a)(s-2b)+(s-2b)(s-2c) +(s-2c)(s-2a) biết S=a+b+c
\(T=\left(b+c-a\right)\left(a+c-b\right)+\left(a+c-b\right)\left(a+b-c\right)+\left(a+b-c\right)\left(b+c-a\right)\)
\(=c^2-\left(a-b\right)^2+a^2-\left(b-c\right)^2+b^2-\left(a-c\right)^2\)
\(=\left(a^2+b^2+c^2\right)-\left(a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\right)\)
\(=2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)\)?????
tính tổng (S-2b)(S-2c)+(S-2c)(S-2a)+(S-2a)(S-2b) trong đó S=a+b+c
\(\text{(S-2b)(S-2c)}=\text{(}a-b+c)\left(a+b-c\right)=a^2-\left(b-c\right)^2=a^2-b^2-c^2+2bc\left(1\right)\)
Hoán vị vòng b->c->a->b, ta được :
\(\text{(S-2c)(S-2a)}=b^2-c^2-a^2+2ca\left(2\right)\)
\(\text{(S-2a)(S-2b)}=c^2-a^2-b^2+2ab\left(3\right)\)
Từ (1),(2),(3) ta suy ra kết quả của tổng là :
\(-a^2-b^2-c^2+2ab+2bc+2ca\)
Tính tổng :
\(\left(S-2b\right)\left(S-2c\right)+\left(S-2c\right)\left(S-2a\right)+\left(S-2a\right)\left(S-2b\right)\)
Trong đó \(S=a+b+c\)
Ta có: \(S=a+b+c\left(1\right)\)
Thay \(\left(1\right)\)vào ta được:
\(\left(S-2b\right).\left(S-2c\right)=\left(a+b+c-2b\right).\)\(\left(a+b+c-2c\right)\)
\(=\left(a-b+c\right).\left(a+b-c\right)\)
\(=a^2+ab-ac-ba-b^2+bc+ca+cb-c^2\)
\(=a^2-b^2-c^2+2.bc\left(2\right)\)
Tương tự, ta được:
\(\left(S-2c\right).\left(S-2a\right)=b^2-c^2-a^2+2.ca\left(3\right)\)
\(\left(S-2a\right).\left(S-2b\right)=c^2-a^2-b^2+2.ab\left(4\right)\)
Từ \(\left(2\right);\left(3\right);\left(4\right)\Rightarrow\)Tổng bằng:
\(a^2-b^2-c^2+2bc+b^2-c^2-a^2+2ca+c^2-a^2\)\(-b^2+2ab\)
\(=2ab+2bc+2ca-a^2-b^2-c^2\)
Vậy tổng trên \(=2ab+2bc+2ca-a^2-b^2-c^2.\)
Thay S=a+b+c vào biểu thức ta được:
(a+b+c-2b)(a+b+c-2c)+(a+b+c-2c)(a+b+c-2a)+(a+b+c-2a)(a+b+c-2b)
=(a-b+c)(a+b-c)+(b-c+a)(b+c-a)+(c-a+b)(c+a-b)
=a2-(b-c)2+b2-(c-a)2+c2-(a-b)2
=a2-b2+2bc-c2+b2-c2+2ac-a2+c2-a2+2ab-b2
=-a2-b2-c2+2ab+2bc+2ca
tinh tong : (S - 2b) (S - 2c) + (S - 2c) (S - 2a) + (S -2a) (S - 2b) trong do S = a+b+c
Chứng minh rằng nếu S = a + b + c thì :
S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b) = (S - 2a)(S - 2b)(S -2c) + 8abc
C/m nếu S = a+b+c thì:
S(S-2b)(S-2c) + S(S-2c)(S-2a) + S(S-2a)(S-2b)
= (S-2a)(S-2b)(S-2c) + 8abc
Chứng minh rằng nếu S = a + b + c thì:
\(S\left(S-2b\right)S\left(S-2c\right)+S\left(S-2c\right)\left(S-2a\right)+S\left(S-2a\right)\left(S-2b\right)=\left(S-2a\right)\left(S-2b\right)\left(S-2c\right)+8abc\)
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
Chứng minh rằng nếu S = a + b + c thì:
\(S\left(S-2b\right)\left(S-2c\right)+S\left(S-2c\right)\left(S-2a\right)+S\left(S-2a\right)\left(S-2b\right)=\left(S-2a\right)\left(S-2b\right)\left(S-2c\right)+8abc\)
Chứng minh rằng nếu S = a + b + c thì:
\(S\left(S-2b\right)\left(S-2c\right)+S\left(S-2c\right)\left(S-2a\right)+S\left(S-2a\right)\left(S-2b\right)=\left(S-2a\right)\left(S-2b\right)\left(S-2c\right)+8abc\)
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