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a: \(=ab\left(a+b\right)-bc\left(b+a\right)-bc\left(c-a\right)-ac\left(c-a\right)\)

\(=\left(a+b\right)\left(ab-bc\right)+\left(a-c\right)\left(bc-ac\right)\)

\(=\left(a+b\right)\cdot b\left(a-c\right)+\left(a-c\right)\cdot c\left(b-a\right)\)

\(=\left(a-c\right)\left(ab+b^2+cb-ac\right)\)

b: \(=ab^2+ac^2+bc^2+a^2b+a^2c+b^2c+2abc\)

\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2\)

\(=\left(a+b\right)\left(ab+c^2+ac+cb\right)\)

\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

d: \(=a^3\left(b-c\right)-b^3\left(b-c+a-b\right)+c^3\left(a-b\right)\)

\(=a^3\left(b-c\right)-b^3\left(b-c\right)-b^3\left(a-b\right)+c^3\left(a-b\right)\)

\(=\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab+b^2-b^2-bc-c^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab-bc-c^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\cdot\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)

 Châu ơi!đăng làm j z

23 tháng 12 2018

1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

15 tháng 8 2018

Bài cuối hơi khó nhìn, bạn thông cảm nhé! ^^

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15 tháng 8 2018

a) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2b-a^2c+c^2a-c^2b+b^2\left(c-a\right)\)

\(=\left(a^2b-c^2b\right)-\left(a^2c-c^2a\right)-b^2\left(a-c\right)\)

\(=b\left(a^2-c^2\right)-ac\left(a-c\right)-b^2\left(a-c\right)\)

\(=b\left(a-c\right)\left(a+c\right)-ac\left(a-c\right)-b^2\left(a-c\right)\)

\(=\left(a-c\right)\left[b\left(a+c\right)-ac-b^2\right]\)

\(=\left(a-c\right)\left(ab+bc-ac-b^2\right)\)

\(=\left(a-c\right)\left[\left(ab-b^2\right)+\left(bc-ac\right)\right]\)

\(=\left(a-c\right)\left[b\left(a-b\right)+c\left(b-a\right)\right]\)

\(=\left(a-c\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]\)

\(=\left(a-c\right)\left(a-b\right)\left(b-c\right)\)

b) \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)

\(=a^3b-a^3c+c^3a-c^3b+b^3\left(c-a\right)\)

\(=\left(a^3b-c^3b\right)-\left(a^3c-c^3a\right)-b^3\left(a-c\right)\)

\(=b\left(a^3-c^3\right)-ac\left(a^2-c^2\right)-b^3\left(a-c\right)\)

\(=b\left(a-c\right)\left(a^2+ac+c^2\right)-ac\left(a-c\right)\left(a+c\right)-b^3\left(a-c\right)\)

\(=\left(a-c\right)\left[b\left(a^2+ac+c^2\right)-ac\left(a+c\right)-b^3\right]\)

\(=\left(a-c\right)\left(ba^2+abc+bc^2-a^2c-ac^2-b^3\right)\)

\(=\left(a-c\right)\left[\left(ba^2-a^2c\right)+\left(abc-ac^2\right)+\left(bc^2-b^3\right)\right]\)

\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)+b\left(c^2-b^2\right)\right]\)

\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)-b\left(b^2-c^2\right)\right]\)

\(=\left(a-c\right)\left[a^2\left(b-c\right)+ac\left(b-c\right)-b\left(b-c\right)\left(b+c\right)\right]\)

\(=\left(a-c\right)\left(b-c\right)\left[a^2+ac-b\left(b+c\right)\right]\)

\(=\left(a-c\right)\left(b-c\right)\left(a^2+ac-b^2-bc\right)\)

\(=\left(a-c\right)\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)

\(=\left(a-c\right)\left(b-c\right)\left(a-b\right)\left(a+b+c\right)\)

1 tháng 6 2018

a) a2(a-b)-b2(a-c)-c2(b-a)

=a2(a-b)-b2(a-c)+c2(a-b)

=(a-b)(a2-c2)-b2(a-c)

=(a-b)(a-c)(a+c)-b2(a-c)

=(a-c)[(a-b)(a+c)-b2]

b)a(b-c)3+b(c-a)3+c(a-b)3

=a(b-c)3-b[(a-b)+(b-c)]+c(a-b)3

=a(b-c)3-b[(a-b)3+3(a-b)2(b-c)+3(a-b)(b-c)2+(b-c)3]+c(a-b)3

=a(b-c)3-b(a-b)3+3b(a-b)2(b-c)+3b(a-b)(b-c)2+b(b-c)3+c(a-b)3

=(b-c)3(a-b)-(a-b)3(b-c)-3b(a-b)(b-c)(a-b+b-c)

=(b-c)3(a-b)-(a-b)3(b-c)-3b(a-b)(b-c)(a-c)

=(a-b)(b-c)[(b-c)2-(a-b)2-3b(a-c)]

=(a-b)(b-c)[(b-c-a+b)(b-c+a-b)-3b(a-c)]

=(a-b)(b-c)[(2b-a-c)(a-c)-3b(a-c)]

=(a-b)(b-c)(a-c)(2b-a-c-3b)

=-(a-b)(b-c)(a-c)(a+b+c)

=(a-b)(b-c)(c-a)(a+b+c)

c)abc-(ab+ac+bc)+(a+b+c)-1

=abc-ab-ac-bc+a+b+c-1

=abc-bc-ab+b-ac+c+a-1

=bc(a-1)-b(a-1)-c(a-1)+a-1

=(a-1)(bc-b-c+1)

=(a-1)[b(c-1)-(c-1)]

=(a-1)(c-1)(b-1)

=(a-1)(b-1)(c-1)

13 tháng 8 2018

a)= ab (a + b) - bc [( a + b) - (a - c)] + ac (a - c)

= ab (a + b) - bc (a + b) + bc (a - c) +ac (a - c)

= b (a + b) (a - c) + c (a - c) (a + b)

= (a + b) (a - c) (b + c)

Y
15 tháng 6 2019

b) \(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left[\left(a^2-b^2\right)+\left(c^2-a^2\right)\right]+\left(c+a\right)\left(c^2-a^2\right)\)

\(=\left(a^2-b^2\right)\left[\left(a+b\right)-\left(b+c\right)\right]+\left(c^2-a^2\right)\left[\left(c+a\right)-\left(b+c\right)\right]\)

\(=\left(a-b\right)\left(a+b\right)\left(a-c\right)-\left(a-c\right)\left(a+c\right)\left(a-b\right)\)

\(=\left(a-b\right)\left(a-c\right)\left[\left(a+b\right)-\left(a+c\right)\right]\)

\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)