Phân tích ab(a+b)-bc(b+c)-ca(c-a)
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LT
0
7 tháng 10 2018
\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(c+a\right)\)
\(=ab\left(a+b+c\right)+bc\left(b+c+a\right)+ca\left(c+a\right)\)
\(=\left(a+b+c\right)\left(ab+bc\right)+ca\left(c+a\right)\)
\(=b.\left(a+b+c\right)\left(a+c\right)+ca\left(c+a\right)\)
\(=\left(a+c\right)\left[b.\left(a+b+c\right)+ca\right]\)
\(=\left(a+c\right)\left(ab+b^2+bc+ca\right)\)
\(=\left(a+c\right)\left[a\left(b+c\right)+b\left(b+c\right)\right]\)
\(=\left(a+c\right)\left(b+c\right)\left(a+b\right)\)
\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc\)
\(=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(c+a\right)+abc\)
\(=ab\left(a+b+c\right)+bc\left(b+c+a\right)+ca\left(c+a+b\right)\)
\(=\left(a+b+c\right)\left(ab+bc+ac\right)\)
Tham khảo nhé~
PT
1
CM
14 tháng 3 2019
Ta có ab(a – b) + bc(b – c) + ca(c – a)
= ab(a – b) + bc[b – a + a – c] + ac(c – a)
= ab(a – b) – bc(a – b) + bc(a – c) – ac(a – c)
= (a – b)(ab – bc) + (a – c)(bc – ac)
= b(a – b)(a – c) – c(a – c)(a – b)
= (a – b)(a – c)(b – c)
Đáp án cần chọn là: A
TP
2
19 tháng 6 2018
\(=ab\left(a-b\right)+b^2c-bc^2+c^2a-ca^2\)
\(=ab\left(a-b\right)+\left(b^2c-ca^2\right)+\left(c^2a-bc^2\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left[\left(ab-bc\right)-\left(ac-c^2\right)\right]\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
19 tháng 6 2018
ab(a-b)+bc(b-c)+ca(c-a)
= ab(a-b)+bc(b-a+a-c)+ca(c-a)
=ab(a-b)+bc(b-a)+bc(a-c)+ca(c-a)
=ab(a-b)-bc(a-b)-bc(c-a)+ca(c-a)
=b(a-b)(a-c)+c(c-a)(a-b)
=(a-b)(a-c)b-c(a-c)(a-b)
=(a-b)(a-c)(b-c)
CM
17 tháng 9 2019
Thêm bớt abc vào M ta có
M = ab(a + b + c) – bc(b + c) – abc + ca(c + a) + abc
= ab(a + b + c) – bc(a + b + c) + ac(a + b + c)
=(a + b + c)(ab – bc + ac)
Đáp án cần chọn là: D
14 tháng 10 2018
a)ab(a+b)-bc(b+c)+ac(a-c)
=ab(a+b)-bc(b+c)+ac\([\left(a+b\right)-\left(b+c\right)]\)
=ab(a+b)-bc(b+c)+ac(a+b)-ac(b+c)
=(a+b)(ab+ac)-(b+c)(bc+ac)
=(a+b)a(b+c)-(b+c)c(b+a)
=(a+b)(b+c)(a-c)
14 tháng 10 2018
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=a^2b+abc+ca^2+ab^2+b^2c+abc+abc+bc^2+ac^2-abc\)
\(=a^2b+a^2c+ab^2+b^2c+c^2a+bc^2+ac^2+2abc\)
\(=\left(a^2b+ba^2+abc\right)+\left(b^2c+c^2b+abc\right)+\left(ac^2+ca^2\right)\)
\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+c\right)\)
\(=\left(a+b+c\right)\left(ab+bc\right)+ac\left(a+c\right)\)
\(=b.\left(a+b+c\right)\left(a+c\right)+ac\left(a+c\right)\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+c\right)\left[b.\left(a+b\right)+c.\left(a+b\right)\right]\)
\(=\left(a+c\right)\left(a+b\right)\left(b+c\right)\)
23 tháng 8 2021
Ta có: \(D=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc\)
\(=a^2b+ab^2+b^2c+bc^2+ac^2+a^2c+3abc\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
24 tháng 8 2021
c: Ta có: \(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a-b\right)^3\cdot\left(a+b\right)\)
BT
3
31 tháng 8 2017
Co P=ab(a-b) + bc((b-a)+(a-c)) +ac(c-a)
=ab(a-b) -bc(a-b) -bc(c-a) +ac(c-a)
=(a-b)(ab-bc) +(c-a)(ac-bc)
=(a-b) b (a-c) + (c-a) c (a-b)
=(a-b)(a-c)(b-c)
\(ab\left(a+b\right)-bc\left(b+c\right)-ca\left(c-a\right)\)
\(=ab\left(a+b\right)-b^2c-bc^2-ac^2+a^2c\)
\(=ab\left(a+b\right)-\left(b^2c-a^2c\right)-\left(bc^2+ac^2\right)\)
\(=ab\left(a+b\right)-c\left(b-a\right)\left(b+a\right)-c^2\left(a+b\right)\)
\(=\left(a+b\right)\left(ab-bc+ac-c^2\right)\)
\(=\left(a+b\right)\text{[}\left(ab+ac\right)-\left(bc+c^2\right)\text{]}\)
\(=\left(a+b\right)\text{[}a\left(b+c\right)-c\left(b+c\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(a-c\right)\)