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\(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab-2bc\)
\(=\left(a+b\right)^2-2.\left(a+b\right).c+c^2-a^2+2ac-c^2-2ab-2bc\)
\(=a^2+2ab+b^2-2ac-2bc+c^2-a^2+2ac-c^2-2ab-2bc\)
\(=b^2-4bc\)
\(=b\left(b-4c\right)\)
(a+b-c)2-(a-c)2-2ab-2bc
=a2+ b2+ c2+ 2ab- 2bc- 2ca- a2+ 2ac- c2- 2ab- 2bc
=b2-4bc
=b.(b-4c)
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 :
\(B=2bc\left(b+2c\right)+2ac\left(c-2a\right)-2ab\left(a+2b\right)-7abc\)
\(B=2b^2c+4bc^2+2ac^2-4a^2c-2ab\left(a+2b\right)-7abc\)
\(B=abc+2b^2c-4a^2c-8abc-2ab\left(a+2b\right)+2ac^2+4bc^2\)
\(B=bc\left(a+2b\right)-4ac\left(a+2b\right)-2ab\left(a+2b\right)+2c^2\left(a+2b\right)\)
\(B=\left(a+2b\right)\left(bc-4ac-2ab+2c^2\right)\)
\(B=\left(a+2b\right)\left[c\left(2c+b\right)-2a\left(2c+b\right)\right]\)
\(B=\left(a+2b\right)\left(2c+b\right)\left(c-2a\right)\)
\(a\left(b^2+c^2\right)+b\left(a^2+c^2\right)+c\left(a^2+b^2\right)+2abc\)
\(=ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+2abc\)
\(=\left(ab^2+ba^2\right)+\left(ac^2+bc^2\right)+\left(ca^2+abc\right)+\left(cb^2+abc\right)\)
\(=ab\left(a+b\right)+c^2\left(a+b\right)+ca\left(a+b\right)+cb\left(a+b\right)\)
\(=\left(a+b\right)\left(ab+c^2+ca+cb\right)\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab-2bc=\left(a+b-c-a+c\right)\left(a+b-c+a-c\right)-2b\left(a+c\right)\)
\(=b\left(2a+b-2c\right)-b\left(2a+2c\right)\)
\(=b\left(2a+b-2c-2a-2c\right)=b\left(b-4c\right)\)