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\(A=\left(a+b+c\right)\left(bc+ac+ab\right)-abc\)
\(=abc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc-abc\)
= \(\left(b^2c+bc^2\right)+\left(a^2c+a^2b\right)+\left(ac^2+abc\right)+\left(ab^2+abc\right)\)
\(=bc\left(b+c\right)+a^2\left(b+c\right)+ac\left(c+b\right)+ab\left(b+c\right)\)
\(=\left(b+c\right)\left(bc+a^2+ac+ab\right)\)
\(=\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
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Ta có : \(A=ab(a-b)+bc(b-c)+ca(c-a)\)
\(\Rightarrow A=ab(a-b)-bc(c-b)+ac(c-a)\)
\(\Rightarrow A=ab(a-b)-bc[(c-a)+(a-b)]+ac(c-a)\)
\(\Rightarrow A=ab(a-b)-bc(a-b)-bc(c-a)+ac(c-a)\)
\(\Rightarrow A=(a-b)(ab-bc)+(c-a)(ac-bc)\)
\(\Rightarrow A=b(a-b)(a-c)-(a-c)c(a-b)\)
\(\Rightarrow A=(a-c)(a-b)(b-c)\)
Chúc học tốt trong kì thi tới :>
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\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)=ab\left(a-b\right)+b^2c-bc^2+c^2a-ca^2\)
\(=ab\left(a-b\right)-\left(ca^2-b^2c\right)+\left(c^2a-bc^2\right)=ab\left(a-b\right)-c\left(a+b\right)\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ca-cb+c^2\right)=\left(a-b\right)\left[\left(ab-ca\right)-\left(cb-c^2\right)\right]\)
\(=\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
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bc(b + c) + ca(c - a) - ab(a + b) = b2c + bc2 + c2a - ca2 - ab(a + b) = (b2c - a2c) + (bc2 + ac2) - ab(a + b)
= c(b - a)(b + a) + c2(b + a) - ab(a + b) = (a + b)[c(b - a) + c2 - ab] = (a + b)[(cb - ab) + (c2 - ca)]
= (a + b)[b(c - a) + c(c - a)] = (a + b)(b + c)(c - a)
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\(C=c\left[b\left(a+d\right)\left(b-c\right)+a\left(b+d\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[\left(ab+bd\right)\left(b-c\right)+\left(ab+ad\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[ab^2-abc+b^2d-bcd+abc-a^2b+acd-a^2d\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[\left(ab^2-a^2b\right)+\left(b^2d-a^2d\right)+\left(acd-bcd\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[ab\left(b-a\right)+d\left(a+b\right)\left(b-a\right)+cd\left(a-b\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left(a-b\right)\left(-ab-da-db+cd\right)+ab\left(c+d\right)\left(a-b\right)\)
\(C=\left(a-b\right)\left(-abc-acd-bcd+c^2d+abc+abd\right)\)
\(C=\left(a-b\right)\left(-acd-bcd+abd+c^2d\right)\)
\(C=c\left(a-b\right)\left(c^2+ab-ac-bc\right)\)
\(C=c\left(a-b\right)\left[\left(c^2-ac\right)-\left(bc-ab\right)\right]\)
\(C=c\left(a-b\right)\left[c\left(c-a\right)-b\left(c-a\right)\right]\)
\(C=c\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
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Phương pháp : xét giá trị riêng
ta có: \(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\) (*)
Nhận thấy vai trò của a,b,c như nhau nên thay a= - b ta đc:
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
=\(\left(-b+b+c\right)\left(-b.b+bc-b.c\right)+b.b+c\)
=\(c.\left(-b^2\right)+b^2.c\)
=0
Suy ra (*) chia hết cho a+b. Mà vai trò của a,b,c như nhau nên (*) chia hết cho (b+c) và (c+a)
=>\(\left(a+b+c\right)\left(ab+bc+ac\right)-abc\)
=\(\left(a+b\right)\left(b+c\right)\left(c+a\right).k\) (**)
ta cho các biến nhận giá trị riêng, chẳng hạn: a=1 ; b=2 ; c=3 thây vào (*) ta đc:
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
=(1+2+3)(1.2+2.3+3.1)-1.2.3=60 (1)
Mặt khác thay a=1 ; b=2 ; c=3 vào (**)ta đc:
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
=(1+2)(2+3)(3+1).k=60.k (2)
từ (1),(2)=> k=60:60=1
Vậy \(\left(a+b+c\right)\left(ab+bc+ca\right)-abc=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
a)
(a+b+c)(ab+bc+ca)-abc
nhân vô rồi rút gọn đươc:
(a^2.b+2abc+c^2.b) +(a^2.c+c^2.a)+(a.b^2+b^2.c)
=b(a+c)^2 +ac(a+c)+ b^2(a+c)
=(a+c)(b(a+c)+ac+b^2)
=(a+c)(ab+b^2+bc+ac)
=(a+c)(b(a+b)+c(a+b))
=(a+c)(a+b)(b+c)