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NV
0
CN
1
KY
15 tháng 9 2019
\(ab\left(a-b\right)-ac\left(a+c\right)+bc\left(2a-b+c\right)\)
\(=a^2b-ab^2-a^2c-ac^2+2abc-b^2c+bc^2\)
\(=a^2b-ab^2-a^2c-ac^2+abc+abc-b^2c+bc^2\)
\(=\left(bc^2-ac^2+abc-a^2c\right)-\left(b^2c-abc-ab^2+a^2b\right)\)
\(=c\left(bc-ac+ab-a^2\right)-b\left(bc-ac-ab+a^2\right)\)
\(=\left(c-b\right)\left(bc-ac+ab-a^2\right)\)
\(=\left(c-b\right)\left[c\left(b-a\right)+a\left(b-a\right)\right]\)
\(=\left(c-b\right)\left(c+a\right)\left(b-a\right)\)
KN
1
NP
0
C
1
MN
2 tháng 3 2020
\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)
\(=ab\left(b-a\right)-b^2c+bc^2-ac^2+a^2c\)
\(=ab\left(b-a\right)+c^2\left(b-a\right)-c\left(b^2-a^2\right)\)
\(=\left(b-a\right)\left(ab+c^2-bc-ca\right)\)
\(=\left(b-a\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(b-a\right)\left(a-c\right)\left(b-c\right)\)
LT
1
C
1
CC
2 tháng 3 2020
\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)
\(=ab\left(b-a\right)-\left(b^2c-bc^2\right)-\left(ac^2-a^2c\right)\)
\(=ab\left(b-a\right)-b^2c+bc^2-ac^2+a^2c\)
\(=ab\left(b-a\right)-\left(b^2c-a^2c\right)+\left(bc^2-ac^2\right)\)
\(=ab\left(b-a\right)-c\left(b^2-a^2\right)+c^2\left(b-a\right)\)
\(=ab\left(b-a\right)-c\left(b-a\right)\left(b+a\right)+c^2\left(b-a\right)\)
\(=\left(b-a\right)\left[ab-c\left(b+a\right)+c^2\right]=\left(b-a\right)\left[ab-\left(bc+ac\right)+c^2\right]\)
\(=\left(b-a\right)\left(ab-bc-ac+c^2\right)=\left(b-a\right)\left[\left(ab-bc\right)-\left(ac-c^2\right)\right]\)
\(=\left(b-a\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]=\left(b-a\right)\left(b-c\right)\left(a-c\right)\)
B
1
TH
2 tháng 3 2020
\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)
\(=ab\left[\left(b-c\right)+\left(c-a\right)\right]-bc\left(b-c\right)-ac\left(c-a\right)\)
\(=ab\left(b-c\right)+ab\left(c-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)
\(=\left[ab\left(b-c\right)-bc\left(b-c\right)\right]+\left[ab\left(c-a\right)-ac\left(c-a\right)\right]\)
\(=\left(b-c\right)\left(ab-bc\right)+\left(c-a\right)\left(ab-ac\right)\)
\(=-b\left(b-c\right)\left(c-a\right)+a\left(c-a\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(c-a\right)\left(a-b\right)\)
NT
2
21 tháng 8 2016
bc(a+d) 9b –c) – ac( b +d) (a-c) + ab(c+d) ( a-b)
= bc(a+d) [ (b-a) + (a-c)] – ac(a-c)(b+d) +ab(c+d)(a-b)
= -bc(a+d )(a-b) +bc(a+d)(a-c) –ac(b+d)(a-c) + ab(c+d)(a-b)
= b(a-b)[ a(c+d) –c(a+d)] + c(a-c)[ b(a+d) –a(b+d)]
= b(a-b). d(a-c) + c(a-c) . d(b-a)
= d(a-b)(a-c)(b-c)
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)