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

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

\(=b\left(2a+b-2c-2a-2c\right)=b\left(b-4c\right)\)

2bc(b + 2c) + 2ac(c - 2a) - 2ab(a + 2b) - 7abc

= 2b²c + 4bc² + 2ac² - 4a²c - 2ab(a + 2b) - 7abc

= 2b²c + abc + 4bc² + 2ac² - 4a²c - 8abc - 2ab(a + 2b)

= bc(2b + a) + 2c²(2b + a) - 4ac(a + 2b) - 2ab(a + 2b)

= (a + 2b)(bc + 2c² - 4ac - 2ab)

= (a + 2b)[c(b + 2c) - 2a(2c + b)]

= (a + 2b)(b + 2c)(c - 2a) 

Lời giải :

\(B=2bc\left(b+2c\right)+2ac\left(c-2a\right)-2ab\left(a+2b\right)-7abc\)

\(B=2b^2c+4bc^2+2ac^2-4a^2c-2ab\left(a+2b\right)-7abc\)

\(B=abc+2b^2c-4a^2c-8abc-2ab\left(a+2b\right)+2ac^2+4bc^2\)

\(B=bc\left(a+2b\right)-4ac\left(a+2b\right)-2ab\left(a+2b\right)+2c^2\left(a+2b\right)\)

\(B=\left(a+2b\right)\left(bc-4ac-2ab+2c^2\right)\)

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

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

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

\(=ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+2abc\)

\(=\left(ab^2+ba^2\right)+\left(ac^2+bc^2\right)+\left(ca^2+abc\right)+\left(cb^2+abc\right)\)

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

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

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

hình như cộng 2abc chứ sao +2ab

Trong đề kt 1t trên lớp mình làm ghi có 2ab à. Vậy chắc là đề sai nhỉ. 

Thanks nha