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\(\left(\text{a}+b+c\right)^3-\text{a}^3-b^3-c^3\)
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29 tháng 9 2019
\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(=ab\left(a-b\right)+bc\left[\left(b-a\right)-\left(c-a\right)\right]+ca\left(c-a\right)\)
\(=ab\left(a-b\right)-bc\left(a-b\right)-bc\left(c-a\right)+ca\left(c-a\right)\)
\(=\left(a-b\right)\left(ab-bc\right)-\left(c-a\right)\left(bc-ca\right)\)
\(=b\left(a-b\right)\left(a-c\right)-c\left(c-a\right)\left(b-a\right)\)
\(=b\left(a-b\right)\left(a-c\right)-c\left(a-c\right)\left(a-b\right)\)
\(=\left(a-c\right)\left(a-b\right)\left(b-c\right)\)
BD
0
HA
2
25 tháng 8 2019
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(a^3-c^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left[\left(a^3-b^3\right)+\left(b^3-c^3\right)\right]+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)
\(=\left(b^3-c^3\right)\left(a-b\right)-\left(a^3-b^3\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(b^2+bc+c^2\right)\left(a-b\right)-\left(a-b\right)\left(a^2+ab+b^2\right)\left(b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(b^2+bc+c^2\right)-\left(a^2+ab+b^2\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(bc+c^2-a^2-ab\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[b\left(c-a\right)+\left(c-a\right)\left(c+a\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
TN
28 tháng 8 2019
a(b3−c3)+b(c3−a3)+c(a3−b3)
=a(b3−c3)−b(a3−c3)+c(a3−b3)
=a(b3−c3)−b[(a3−b3)+(b3−c3)]+c(a3−b3)
=a(b3−c3)−b(b3−c3)−b(a3−b3)+c(a3−b3)
=(b3−c3)(a−b)−(a3−b3)(b−c)
=(b−c)(b2+bc+c2)(a−b)−(a−b)(a2+ab+b2)(b−c)
=(a−b)(b−c)[(b2+bc+c2)−(a2+ab+b2)]
=(a−b)(b−c)(bc+c2−a2−ab)
=(a−b)(b−c)[b(c−a)+(c−a)(c+a)]
=(a−b)(b−c)(c−a)(a+b+c)
HA
0
TD
1
QT
Quoc Tran Anh Le
Giáo viên
1 tháng 7 2019
\(\left(a+b+c\right)^2+\left(a+b-c\right)^2-4c^2\)
\(=\left[\left(a+b+c\right)^2-\left(2c\right)^2\right]+\left(a+b-c\right)^2\)
\(=\left(a+b+3c\right)\left(a+b-c\right)+\left(a+b-c\right)^2\)
\(=\left(a+b-c\right)\left(a+b+3c+a+b-c\right)\)
\(=\left(a+b-c\right)\left(2a+2b+2c\right)\)
\(=2\left(a+b-c\right)\left(a+b+c\right)\)
LA
1
1 tháng 7 2019
\(=\left(a+b\right)^2+2\left(a+b\right)c+c^2+\left(a+b\right)^2-2\left(a+b\right)c+c^2-4c^2\)
\(=2\left(a+b\right)^2-2c^2=2\left[\left(a+b\right)^2-c^2\right]=2\left(a+b+c\right)\left(a+b-c\right)\)
VN
2
QT
Quoc Tran Anh Le
Giáo viên
1 tháng 7 2019
Mình đã làm bài này rồi.
Link: https://hoc24.vn/hoi-dap/question/824554.html
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3a^2b+3ab^2+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+2abc+b^2c+bc^2\right)\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+abc+abc+b^2c+bc^2\right)\)
\(=3\left[ab\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+c^2+ac+bc\right)\)
\(=3\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)