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(a + b + c)(ab + bc + ac) - abc
= a2b + abc +a2c + ab2 + b2c + abc + abc + bc2 + ac2
= (a2b + 2abc + bc2) + (ac2 + a2c) + (ab2 + b2c)
= b(a2 + 2ac + c2) + ac(c + a) + b2(a + c)
= b(a + c) + ac(a + c) + b2(a + c)
= (a + c)[b(a + c) + ac + b2]
= (a + c)(ab + bc + ac + b2)
= (a + c)[b(a + b) + c(a + b)]
= (a + c)(b + c)(a + b)
\(\left(a+b+c\right)\left(ab+bc+ac\right)-abc\)
\(=a^2b+abc+a^2c+b^2a+b^2c+abc+abc+c^2b+c^2a-abc\)
\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a^2+b^2+2ab-2ab\right)+2abc\)
\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2-2abc+2abc\)
\(=\left(a+b\right)\left(ab+c^2+ca+cb\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
sửa đề thành \(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=ab\left(a+b\right)+b^2c+bc^2+c^2a+ca^2+2abc\)
\(=ab\left(a+b\right)+\left(b^2c+abc\right)+\left(c^2a+c^2b\right)+\left(a^2c+abc\right)\)
\(=ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\)
\(=\left(a+b\right)\left(ab+bc+a^2+ca\right)\)
\(=\left(a+b\right)\left[\left(ab+bc\right)+\left(c^2+ac\right)\right]\)
\(=\left(a+b\right)\left[b\left(a+c\right)+c\left(c+a\right)\right]\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(ab\left(a-b\right)-ac\left(a+c\right)+bc\left(2a-b+c\right)\)
\(=a^2b-ab^2-a^2c-ac^2+2abc-b^2c+bc^2\)
\(=a^2b-ab^2-a^2c-ac^2+abc+abc-b^2c+bc^2\)
\(=\left(bc^2-ac^2+abc-a^2c\right)-\left(b^2c-abc-ab^2+a^2b\right)\)
\(=c\left(bc-ac+ab-a^2\right)-b\left(bc-ac-ab+a^2\right)\)
\(=\left(c-b\right)\left(bc-ac+ab-a^2\right)\)
\(=\left(c-b\right)\left[c\left(b-a\right)+a\left(b-a\right)\right]\)
\(=\left(c-b\right)\left(c+a\right)\left(b-a\right)\)
\(ab\left(a+b\right)-bc\left(b+c\right)+ca\left(a+c\right)+abc\)
\(=a^2b+ab^2-b^2c-bc^2+ca^2+c^2b+abc\)
\(=a^2b+ab^2-b^2c+a^2c+abc\)
Đến đây thì mk chịu
https://h7.net/hoi-dap/toan-8/phan-h-da-thuc-abc-ab-bc-ca-a-b-c-1-thanh-nhan-tu-faq382483.html