phân tích nhân tử ab(a-b)+bc(b-c)+ca(c-a)
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DD
2
14 tháng 10 2018
a)ab(a+b)-bc(b+c)+ac(a-c)
=ab(a+b)-bc(b+c)+ac\([\left(a+b\right)-\left(b+c\right)]\)
=ab(a+b)-bc(b+c)+ac(a+b)-ac(b+c)
=(a+b)(ab+ac)-(b+c)(bc+ac)
=(a+b)a(b+c)-(b+c)c(b+a)
=(a+b)(b+c)(a-c)
14 tháng 10 2018
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=a^2b+abc+ca^2+ab^2+b^2c+abc+abc+bc^2+ac^2-abc\)
\(=a^2b+a^2c+ab^2+b^2c+c^2a+bc^2+ac^2+2abc\)
\(=\left(a^2b+ba^2+abc\right)+\left(b^2c+c^2b+abc\right)+\left(ac^2+ca^2\right)\)
\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+c\right)\)
\(=\left(a+b+c\right)\left(ab+bc\right)+ac\left(a+c\right)\)
\(=b.\left(a+b+c\right)\left(a+c\right)+ac\left(a+c\right)\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+c\right)\left[b.\left(a+b\right)+c.\left(a+b\right)\right]\)
\(=\left(a+c\right)\left(a+b\right)\left(b+c\right)\)
BT
3
31 tháng 8 2017
Co P=ab(a-b) + bc((b-a)+(a-c)) +ac(c-a)
=ab(a-b) -bc(a-b) -bc(c-a) +ac(c-a)
=(a-b)(ab-bc) +(c-a)(ac-bc)
=(a-b) b (a-c) + (c-a) c (a-b)
=(a-b)(a-c)(b-c)
D
1
VT
2 tháng 9 2017
sửa đề thành \(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=ab\left(a+b\right)+b^2c+bc^2+c^2a+ca^2+2abc\)
\(=ab\left(a+b\right)+\left(b^2c+abc\right)+\left(c^2a+c^2b\right)+\left(a^2c+abc\right)\)
\(=ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\)
\(=\left(a+b\right)\left(ab+bc+a^2+ca\right)\)
\(=\left(a+b\right)\left[\left(ab+bc\right)+\left(c^2+ac\right)\right]\)
\(=\left(a+b\right)\left[b\left(a+c\right)+c\left(c+a\right)\right]\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
7 tháng 10 2018
\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(c+a\right)\)
\(=ab\left(a+b+c\right)+bc\left(b+c+a\right)+ca\left(c+a\right)\)
\(=\left(a+b+c\right)\left(ab+bc\right)+ca\left(c+a\right)\)
\(=b.\left(a+b+c\right)\left(a+c\right)+ca\left(c+a\right)\)
\(=\left(a+c\right)\left[b.\left(a+b+c\right)+ca\right]\)
\(=\left(a+c\right)\left(ab+b^2+bc+ca\right)\)
\(=\left(a+c\right)\left[a\left(b+c\right)+b\left(b+c\right)\right]\)
\(=\left(a+c\right)\left(b+c\right)\left(a+b\right)\)
\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc\)
\(=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(c+a\right)+abc\)
\(=ab\left(a+b+c\right)+bc\left(b+c+a\right)+ca\left(c+a+b\right)\)
\(=\left(a+b+c\right)\left(ab+bc+ac\right)\)
Tham khảo nhé~
NT
4
1 tháng 8 2015
X=ab(a+b)+bc(b+c)+ca(c+a)+2abc
=a2b+ab2+b2c+bc2+c2a+a2c+2abc
=b(a+c)2+b2(a+c)+ac(a+c)+2abc
=[b(a+c)+b2+ac](a+c)+2abc
=(ab+bc+b2+ac)(a+c)+2abc
=[b(c+b)+a(b+c)](a+c)+2abc
=[(b+a)(b+c)](a+c)+2abc
rồi j nữa, tịt rồi
TT
1
16 tháng 10 2015
bc(b+c)+ca(c-a)-ab(a+b)
=bc(b+c)+ca(b+c)-ca(a+b)-ab(a+b)
=c(a+b)(b+c)-a(a+b)(b+c)
=(c-a)(a+b)(b+c)
LP
0
DT
1
6 tháng 5 2016
\(T=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+abc+abc\)
\(T=a^2b+ab^2+abc+b^2c+bc^2+abc+c^2a+a^2c\)
\(T=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+c\right)\)
\(T=\left(a+b+c\right)\left(b\left(a+c\right)\right)+ac\left(a+c\right)\)
\(T=\left(a+c\right)\left(b\left(a+b+c\right)+ac\right)\)
NL
1
EC
3 tháng 9 2016
\(ab\left(a+b\right)-bc\left(b+c\right)+ca\left(a+c\right)+abc\)
\(=a^2b+ab^2-b^2c-bc^2+ca^2+c^2b+abc\)
\(=a^2b+ab^2-b^2c+a^2c+abc\)
Đến đây thì mk chịu
\(=ab\left(a-b\right)+b^2c-bc^2+c^2a-ca^2\)
\(=ab\left(a-b\right)+\left(b^2c-ca^2\right)+\left(c^2a-bc^2\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left[\left(ab-bc\right)-\left(ac-c^2\right)\right]\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
ab(a-b)+bc(b-c)+ca(c-a)
= ab(a-b)+bc(b-a+a-c)+ca(c-a)
=ab(a-b)+bc(b-a)+bc(a-c)+ca(c-a)
=ab(a-b)-bc(a-b)-bc(c-a)+ca(c-a)
=b(a-b)(a-c)+c(c-a)(a-b)
=(a-b)(a-c)b-c(a-c)(a-b)
=(a-b)(a-c)(b-c)