ab(a+b)+bc(b+c)+ca(a+c)+3abc
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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 \(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=ab\left(a+b\right)+b^2c+bc^2+c^2a+ca^2+2abc\)
\(=ab\left(a+b\right)+\left(b^2c+abc\right)+\left(c^2a+c^2b\right)+\left(a^2c+abc\right)\)
\(=ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\)
\(=\left(a+b\right)\left(ab+bc+a^2+ca\right)\)
\(=\left(a+b\right)\left[\left(ab+bc\right)+\left(c^2+ac\right)\right]\)
\(=\left(a+b\right)\left[b\left(a+c\right)+c\left(c+a\right)\right]\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
BC
2
7 tháng 3 2021
\(A=\left(a+b+c\right).\left(bc+ca+ab\right)-abc\\ =abc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc-abc\\ =\left(b^2c+bc^2\right)+\left(a^2b+a^2c\right)+\left(ac^2+abc\right)+\left(ab^2+abc\right)\\ =bc\left(b+c\right)+a^2\left(b+c\right)+ac\left(b+c\right)+ab\left(b+c\right)\\ =\left(b+c\right)\left(bc+a^2+ac+ab\right)\\ =\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]=\left(b+c\right)\left(a+c\right)\left(a+b\right)\)
LP
0
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
KQ
1
7 tháng 10 2018
ab(a+b) + bc(b+c) + ca(c+a) = a^2b + ab^2 + b^2c + bc^2 + ca(c+a) + 2abc
= ab^2 + b^2c + a^2b + bc^2 + 2abc + ca(c+a)
=b^2(a+c) + b(a^2 + c^2 + 2ac) + ca(c+a)
=b^2(a+c) + b(a+c)^2 + ca(c+a)
=(c+a)[b^2 + b(a+c) + ca]
=(c+a)[b^2 + ab + bc + ca]
=(c+a)[b(b+a) + c(b+a)]
=(c+a)(b+c)(b+a)
NG
1
19 tháng 9 2021
\(=bc\left(b+c\right)\)\(+ac^2\)\(-a^2\)\(c\)\(-a^2\)\(b\)\(-ab^2\)
\(=bc\left(b+c\right)\)\(-a^2\)\(\left(b+c\right)\)\(-a\left(b+c\right)\)\(\left(b-c\right)\)
\(=\left(b+c\right)\)\(\left(bc-a^2-ab+ac\right)\)
\(=\left(b+c\right)\)\(\orbr{c\left(b+a\right)-a\left(a+b\right)}\)
\(=\left(b+c\right)\)\(\left(b+a\right)\)\(\left(c-a\right)\)
Tỉ lệ sai 1%
NH
5
PL
26 tháng 2 2020
(a+b+c)(ab+bc+ca)−abc
=(a+b)(ab+bc+ac)+c(ab+bc+ca)−abc
=(a+b)(ab+bc+ca)+abc+c2(a+b)−abc
=(a+b)(ab+bc+ca+c2)
=(a+b)(b+c)(c+a)
nguồn: https://h7.net/hoi-dap/toan-8/phan-h-a-b-c-ab-bc-ca-abc-thanh-nhan-tu--faq429360.html
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~
ta có: ab(a + b) + bc(b + c) + ac(a + c) + 3abc
= ab(a + b) + abc + bc(b + c) + abc + ac(a + c) + abc
= ab(a + b + c) + bc(a + b + c) + ac(a + b + c)
= (a + b + c)(ab + bc + ca)