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\(A=\left(a+b+c\right)\left(bc+ac+ab\right)-abc\)
\(=abc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc-abc\)
= \(\left(b^2c+bc^2\right)+\left(a^2c+a^2b\right)+\left(ac^2+abc\right)+\left(ab^2+abc\right)\)
\(=bc\left(b+c\right)+a^2\left(b+c\right)+ac\left(c+b\right)+ab\left(b+c\right)\)
\(=\left(b+c\right)\left(bc+a^2+ac+ab\right)\)
\(=\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 :>
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)=ab\left(a-b\right)+b^2c-bc^2+c^2a-ca^2\)
\(=ab\left(a-b\right)-\left(ca^2-b^2c\right)+\left(c^2a-bc^2\right)=ab\left(a-b\right)-c\left(a+b\right)\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ca-cb+c^2\right)=\left(a-b\right)\left[\left(ab-ca\right)-\left(cb-c^2\right)\right]\)
\(=\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=c\left[b\left(a+d\right)\left(b-c\right)+a\left(b+d\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[\left(ab+bd\right)\left(b-c\right)+\left(ab+ad\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[ab^2-abc+b^2d-bcd+abc-a^2b+acd-a^2d\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[\left(ab^2-a^2b\right)+\left(b^2d-a^2d\right)+\left(acd-bcd\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[ab\left(b-a\right)+d\left(a+b\right)\left(b-a\right)+cd\left(a-b\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left(a-b\right)\left(-ab-da-db+cd\right)+ab\left(c+d\right)\left(a-b\right)\)
\(C=\left(a-b\right)\left(-abc-acd-bcd+c^2d+abc+abd\right)\)
\(C=\left(a-b\right)\left(-acd-bcd+abd+c^2d\right)\)
\(C=c\left(a-b\right)\left(c^2+ab-ac-bc\right)\)
\(C=c\left(a-b\right)\left[\left(c^2-ac\right)-\left(bc-ab\right)\right]\)
\(C=c\left(a-b\right)\left[c\left(c-a\right)-b\left(c-a\right)\right]\)
\(C=c\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a b<a+b> <a-b> + bc < b - c> < b + c >+ ca < c - a > < c + a>
a² b+ ab² + a² b - ab² + b² c -bc² +b² c + bc² + c² a -ca² + c² a +ca²
<a² b +a² b> + < ab² - ab² > + < b²c + b² c > + <-bc² + bc² > + < c² a +c² a> + <-ca² + ca² >
2 a² b + 2 b² c +2 c² a
XONG NHA NGƯỜI ANH EM
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=bc\left(a+d\right)\left(b-c\right)-ac\left(b+d\right)\left(a-c\right)+ab\left(c+d\right)\left(a-b\right)\)
\(=bc\left(a+d\right)\left[\left(b-a\right)+\left(a-c\right)\right]-ac\left(a-c\right)\left(b+d\right)+ab\left(c+d\right)\)\(\left(a-b\right)\)
\(=bc\left(a+d\right)\left(a-b\right)+bc\left(a+d\right)\left(a-c\right)-ac\left(b+d\right)\left(a-c\right)\)\(+ab\left(c+d\right)\left(a-b\right)\)
\(=b\left(a-b\right)\left[a\left(c+d\right)-c\left(a+d\right)\right]+c\left(a-c\right)\left[b\left(a+d\right)-a\left(b+d\right)\right]\)
\(=b\left(a-b\right).d\left(a-c\right)+c\left(a-c\right).d\left(b-a\right)\)
\(=d\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
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)