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)

=d(a-b)(c-a)(c-b)

A= bc(a+d)(b-c) +ac(b+d)(c-a) + ab(c+d)(a-b) 
A= bc(ab+ bd -ac -dc ) + ac(bc+cd -ab-ad )+ab(ac+ad-bc-bd) 
A=(ab²c + b²cd -abc² -bdc² ) + (abc² + adc² -a²bc -a²cd ) + (a²bc + a²bd - ab²c -ab²d) 
A= (ab²c + cb²d -ab²c-ab²d) + (c²ab -abc² -bdc² +adc² ) + ( a²bd +a²bc -a²bc -a²cd) 
A= a²(bd-cd) + b²(cd-ad) + c²(ad-bd) 
A=a²d(b-c) + b²d(c-a) + c²d(a-b) 
A=d(a²b-a²c + b²c-b²a +c²a-c²b) 
A=d[b(a²-c²) + c(b²-a²) + a(c² - b²)] 

gimf mk nha

Tacó a-c = a-c+b+b = (a-b)(b-c)

A=bc(a+d)(b-c)-ac(b+d)[(a-b)+(b-c)]+ab(c+d)(a-b)

A=bc(a+d)(b-c)-ac(b+d)(a-b)-ac(b+d)(b-c)+ab(c+d)(a-b)

A=(b-c)[bc(a+d)-ac(b+d)]+(a-b)[-ac(b+d)+ab(c+a)]

A=(b-c)(abc+bcd-abc-acd)+(a-b)(-bca-acd+abc+abd)

A=(b-c)cd(b-a)+(a-b)ad(b-c)

A=(b-c)(b-a)(cd-ad)

A=(b-c)(b-a)d(c-a)=d(b-c)(b-a)(c-a)

\(C=c\left[b\left(a+d\right)\left(b-c\right)+a\left(b+d\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)

\(C=c\left[\left(ab+bd\right)\left(b-c\right)+\left(ab+ad\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)

\(C=c\left[ab^2-abc+b^2d-bcd+abc-a^2b+acd-a^2d\right]+ab\left(c+d\right)\left(a-b\right)\)

\(C=c\left[\left(ab^2-a^2b\right)+\left(b^2d-a^2d\right)+\left(acd-bcd\right)\right]+ab\left(c+d\right)\left(a-b\right)\)

\(C=c\left[ab\left(b-a\right)+d\left(a+b\right)\left(b-a\right)+cd\left(a-b\right)\right]+ab\left(c+d\right)\left(a-b\right)\)

\(C=c\left(a-b\right)\left(-ab-da-db+cd\right)+ab\left(c+d\right)\left(a-b\right)\)

\(C=\left(a-b\right)\left(-abc-acd-bcd+c^2d+abc+abd\right)\)

\(C=\left(a-b\right)\left(-acd-bcd+abd+c^2d\right)\)

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

\(C=c\left(a-b\right)\left[\left(c^2-ac\right)-\left(bc-ab\right)\right]\)

\(C=c\left(a-b\right)\left[c\left(c-a\right)-b\left(c-a\right)\right]\)

\(C=c\left(a-b\right)\left(c-a\right)\left(c-b\right)\)

   bc(a+d)(b-c)-ac(b+d)(a-c)+ab(c+d)(a-b)

=bc(ab+bd-ac-cd)-ac(ab+ad-bc-cd)+ab(ac+ad-bc-bd)

=ab²c+cb²d-abc²-bc²d-a²bc-a²cd+abc²+ac²d+a²bc+a²bd-ab²c-ab²d

=b²cd-bc²d-a²cd+ac²d+a²bd-ab²d

=bcd(b-c)+a²d(b-c)-ad(b²-c²)

=bcd(b-c)+a²d(b-c)-ad(b-c)(b+c)

=(b-c)[bcd+a²d-ad(b+c)]

=(b-c)(bcd+a²d-abd-acd)

=(b-c)[ad(a-c)-bd(a-c)]

=(b-c)(a-c)(ad-bd)

=(b-c)(a-c)(a-b)d

Ta có:

\(A=bc\left(a+d\right)\left(b-c\right)-ac\left(b+d\right)\left(a-c\right)+ab\left(c+d\right)\left(a-b\right)\)

\(=bc\left(a+d\right)\left[\left(b-a\right)+\left(a-c\right)\right]-ac\left(a-c\right)\left(b+d\right)+ab\left(c+d\right)\)\(\left(a-b\right)\)

\(=bc\left(a+d\right)\left(a-b\right)+bc\left(a+d\right)\left(a-c\right)-ac\left(b+d\right)\left(a-c\right)\)\(+ab\left(c+d\right)\left(a-b\right)\)

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

\(=b\left(a-b\right).d\left(a-c\right)+c\left(a-c\right).d\left(b-a\right)\)

\(=d\left(a-b\right)\left(a-c\right)\left(b-c\right)\)

a:

\(a^3+a^2c-abc+b^2c+b^3\)

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

\(=\left(a^2-ab+b^2\right)\left(a+b+c\right)=0\)(vì a+b=c=0)

câu b bn xem ở link này nha!

Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath

\(a^3+a^2c-abc+b^2c+b^3\)

\(=\left(a^3+b^3\right)\left(a^2c-abc+b^2c\right)\)

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

\(\Rightarrow\left(a+b+c\right)\left(a^2-ab+b^2\right)=0\)( vì a+b+c=0)

Vậy \(a^3+a^2c-abc+b^2c+b^3=0\left(đpcm\right)\)

\(b,A=bc\left(a+d\right)\left(b-c\right)-ac\left(b+d\right)\left(a-c\right)+ab\left(c+d\right)\left(a-b\right)\)

\(=bc\left(a+d\right)\left[\left(b-a\right)+\left(a-c\right)\right]-ac\left(a-c\right)\left(b+d\right)+ab\left(c+d\right)\left(a-b\right)\)

\(=-bc\left(a+d\right)\left(a-b\right)+bc\left(a+d\right)\left(a-c\right)-ac\left(a-c\right)\left(b+d\right)+ab\left(c+d\right)\left(a-b\right)\)

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

\(=b\left(a-b\right)\cdot d\left(a-c\right)+c\left(a-c\right)\cdot d\left(b-a\right)\)

\(=d\left(a-b\right)\left(a-c\right)\left(b-c\right)\)

d) (b+c)(b+a)(c-a)

c) (b-1)(ac+1-a-c)

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