Phân tích thành nhân tử \(\left(a+b\right)\left(b-c\right)\left(c+a\right)+abc\)
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27 tháng 12 2017
A = a^2b+ab^2+b^2c+bc^2+c^2a+ca^2+3abc
= (a^2b+ab^2+abc)+(b^2c+bc^2+abc)+(c^2a+ca^2+abc)
= (a+b+c).(ab+bc+ca)
k mk nha
27 tháng 12 2017
( a + b ) ( b + c ) ( c + a ) + abc
= ( ab + ac + b^2 + bc ) ( c + a ) + abc
= ( c + a ) ( ab + ac + b^2 + bc ) + abc
= abc + ac^2 + b^2c + bc^2 + a^2b + a^2c + ab^2 + abc + abc
= 3abc + a . c^2 + b^2 . c + b . c^2 + a^2 . b + a^2 . c + a . b^2
= 3abc + c^2( a + b ) + b^2( c + a ) + a^2 ( b + c )
21 tháng 9 2016
a3(c - b2) + b3(a - c2) + c3(b - a2) + abc(abc - 1)
= a3c - a3b2 + ab3 - b3c2 + bc3 - a2c3 + a2b2c2 - abc
= a2b2c2 - b3c2 - (a2c3 - bc3) - (a3b2 - ab3) + (a3c - abc)
= b2c2(a2 - b) - c3(a2 - b) - ab2(a2 - b) + ac(a2 - b)
= (a2 - b)(b2c2 - c3 - ab2 + ac) = (a2 - b)[c2(b2 - c) - a(b2 - c)] = (a2 - b)(b2 - c)(c2 - a)
TT
10 tháng 12 2020
\(A=\left(a+b+c\right)\left(bc+ac+ab\right)-abc\)
\(=abc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc-abc\)
= \(\left(b^2c+bc^2\right)+\left(a^2c+a^2b\right)+\left(ac^2+abc\right)+\left(ab^2+abc\right)\)
\(=bc\left(b+c\right)+a^2\left(b+c\right)+ac\left(c+b\right)+ab\left(b+c\right)\)
\(=\left(b+c\right)\left(bc+a^2+ac+ab\right)\)
\(=\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
29 tháng 9 2018
a3(c - b2) + b(a - c2) + c3(b - a2) + abc(abc - 1)
= a3c - a3b2 + ab3 - b3c2 + c3b - a2c3 + a2b2c2 - abc
= (a2b2c2 - b3c2) + (a3c - abc) - (a3b2 - ab3) - (a2c3 - c3b)
= b2c2(a2 - b) + ac(a2 - b) - ab2(a2 - b) - c3(a2 - b)
= (a2 - b)(b2c2 + ac - ab2 - c3)
= (a2 - b)[(b2c2 - c3) - (ab2 - ac)]
= (a2 - b)[c2(b2 - c) - a(b2 - c)]
= (a2 - b)(c2 - a)(b2 - c)
5 tháng 10 2023
\(C=c\left[b\left(a+d\right)\left(b-c\right)+a\left(b+d\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[\left(ab+bd\right)\left(b-c\right)+\left(ab+ad\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[ab^2-abc+b^2d-bcd+abc-a^2b+acd-a^2d\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[\left(ab^2-a^2b\right)+\left(b^2d-a^2d\right)+\left(acd-bcd\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[ab\left(b-a\right)+d\left(a+b\right)\left(b-a\right)+cd\left(a-b\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left(a-b\right)\left(-ab-da-db+cd\right)+ab\left(c+d\right)\left(a-b\right)\)
\(C=\left(a-b\right)\left(-abc-acd-bcd+c^2d+abc+abd\right)\)
\(C=\left(a-b\right)\left(-acd-bcd+abd+c^2d\right)\)
\(C=c\left(a-b\right)\left(c^2+ab-ac-bc\right)\)
\(C=c\left(a-b\right)\left[\left(c^2-ac\right)-\left(bc-ab\right)\right]\)
\(C=c\left(a-b\right)\left[c\left(c-a\right)-b\left(c-a\right)\right]\)
\(C=c\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
11 tháng 4 2018
\(=a\left(b+c\right)\left(b^2-c^2\right)+b\left(c+a\right)\left(c^2-a^2\right)+c\left(a+b\right)\left(a^2-b^2\right)\)
\(=\left(ab+ac\right)\left(b^2-c^2\right)+\left(bc+ba\right)\left(c^2-a^2\right)+\left(ca+cb\right)\left(a^2-b^2\right)\)
\(=ab^3+ab^2c-abc^2-ac^3+bc^3+abc^2-a^2bc-a^3b+a^3c+a^2bc-ab^2c-b^3c\)
\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)
\(=\left(ab^3-b^3c\right)+\left(bc^3-ac^3\right)+\left(a^3c-a^3b\right)\)
\(=b^3\left(a-c\right)+c^3\left(b-a\right)+a^3\left(c-b\right)\)
\(=b^3\left(a-c\right)+c^3\left(c-a+b-c\right)+a^3\left(c-b\right)\)(Đổi dấu hạng tử ở giữa)
\(=b^3\left(a-c\right)-c^3\left(a-c\right)-c^3\left(c-b\right)+a^3\left(c-b\right)\)
\(=\left(a-c\right)\left(b^3-c^3\right)-\left(b-c\right)\left(a^3-c^3\right)\)
\(=\left(a-c\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(a-c\right)\left(b-c\right)\left(a^2+ac+c^2\right)\)
\(=\left(a-c\right)\left(b-c\right)\left(b^2+bc+c^2-a^2-ac-c^2\right)\)
\(=\left(a-c\right)\left(b-c\right)\left(b^2-a^2-ac+bc\right)\)
\(=\left(a-c\right)\left(b-c\right)[\left(b-a\right)\left(b+a\right)+c\left(b-a\right)]\)
\(=\left(a-c\right)\left(b-c\right)\left(b-a\right)\left(a+b+c\right)\)
Sửa:\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=\left(ab+ac+b^2+bc\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ac^2+a^2c+b^2c+b^2a+bc^2+abc+abc\)
\(=ab\left(a+b\right)+bc\left(b+c\right)+ac\left(c+a\right)+abc+abc+abc\)
\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)