Phân tích đa thức sau thành nhân tử:
\(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)
Mình đang cần bài gấp,
Trưa đi học rồi nha !
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26 tháng 9 2016
(a+b+c)^3 thì viết được thành [(a+b)+c)]^3 rồi AD hằng đẳng thức để tính. Còn với (a^3+b^3+c^3) ta viết được (a+b)^3 -3a^2b -3ab^2 + c^3=(a+b)^3 -3ab(a+b)+c^3 ...thay vào rồi đổi biến
16 tháng 11 2016
a(b3-c3) -b(b3-c3+a3-b3)+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+cb+c2)(a-b)-(a-b)(a2+ab+b2)(b-c)
=(b-c)(a-b)(b2+Cb+c2-a2-ab-b2)
=(b-c)(a-b)(c2+cb-ab-a2)
=(b-c)(a-b)[(c-a)(c+a)+b(c-a)]
=(b-c)(a-b)(c-a)(a+c+b)
26 tháng 9 2017
\(a^4\left(b^2-c^2\right)+b^4\left(c^2-a^2\right)+c^4\left(a^2-b^2\right)\)
\(=a^4\left(b^2-c^2\right)+b^4\left(c^2-b^2+b^2-a^2\right)+c^4\left(a^2-b^2\right)\)
\(=a^4\left(b^2-c^2\right)+b^4\left(c^2-b^2\right)+b^4\left(b^2-a^2\right)+c^4\left(a^2-b^2\right)\)
\(=a^4\left(b^2-c^2\right)-b^4\left(b^2-c^2\right)-b^4\left(a^2-b^2\right)+c^4\left(a^2-b^2\right)\)
\(=\left(a^4-b^4\right)\left(b^2-c^2\right)+\left(c^4-b^4\right)\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)\left(b^2-c^2\right)-\left(b^2-c^2\right)\left(c^2+b^2\right)\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(b^2-c^2\right)\left(a^2+b^2-c^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(b^2-c^2\right)\left(a^2-c^2\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(b-c\right)\left(b+c\right)\left(a-c\right)\left(a+c\right)\)
1 tháng 11 2018
\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2-a^3-b^3-c^3+4abc\)
\(=a\left(b-c\right)^2-a^3+4abc+b\left(c-a\right)^2-b^3+c\left(a-b\right)^2-c^3\)
\(=a\left[\left(b-c\right)^2+4bc-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)
\(=a\left[\left(b+c\right)^2-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)
\(=a\left(b+c+a\right)\left(b+c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[a\left(b+c+a\right)+b\left(c-a-b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[ab+ac+a^2+bc-ab-b^2\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[c\left(a+b\right)+\left(a-b\right)\left(a+b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left(a+b\right)\left(a-b+c\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(a-b+c\right)\left[b^2-\left(a-c\right)^2\right]\)
\(=\left(a-b+c\right)\left(b+a-c\right)\left(b-a+c\right)\)
5 tháng 10 2023
\(C=c\left[b\left(a+d\right)\left(b-c\right)+a\left(b+d\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[\left(ab+bd\right)\left(b-c\right)+\left(ab+ad\right)\left(c-a\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[ab^2-abc+b^2d-bcd+abc-a^2b+acd-a^2d\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[\left(ab^2-a^2b\right)+\left(b^2d-a^2d\right)+\left(acd-bcd\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left[ab\left(b-a\right)+d\left(a+b\right)\left(b-a\right)+cd\left(a-b\right)\right]+ab\left(c+d\right)\left(a-b\right)\)
\(C=c\left(a-b\right)\left(-ab-da-db+cd\right)+ab\left(c+d\right)\left(a-b\right)\)
\(C=\left(a-b\right)\left(-abc-acd-bcd+c^2d+abc+abd\right)\)
\(C=\left(a-b\right)\left(-acd-bcd+abd+c^2d\right)\)
\(C=c\left(a-b\right)\left(c^2+ab-ac-bc\right)\)
\(C=c\left(a-b\right)\left[\left(c^2-ac\right)-\left(bc-ab\right)\right]\)
\(C=c\left(a-b\right)\left[c\left(c-a\right)-b\left(c-a\right)\right]\)
\(C=c\left(a-b\right)\left(c-a\right)\left(c-b\right)\)
\(a^4\left(b-c\right)+b^4\left(c-a\right)+c^4\left(a-b\right)\)
\(=a^4\left(b-c\right)+b^4[\left(c-b\right)-\left(a-b\right)]+c^4\left(a-b\right)\)
\(=a^4\left(b-c\right)+b^4\left(c-b\right)-b^4\left(a-b\right)+c^4\left(a-b\right)\)
\(=a^4\left(b-c\right)-b^4\left(b-c\right)-b^4\left(a-b\right)+c^4\left(a-b\right)\)
\(=\left(b-c\right)\left(a^4-b^4\right)-\left(a-b\right)\left(c^4-b^4\right)\)
\(=\left(b-c\right)\left(a^2-b^2\right)\left(a^2+b^2\right)-\left(a-b\right)\left(c^2-b^2\right)\left(c^2+b^2\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)+\left(a-b\right)\left(b-c\right)\left(c+b\right)\left(c^2+b^2\right)\)
\(=\left(b-c\right)\left(a-b\right)[\left(a+b\right)\left(a^2+b^2\right)+\left(c+b\right)\left(c^2+b^2\right)]\)