Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: Sửa đề: \(a^2\left(a+1\right)+b^2\left(b-1\right)-a^2b^2\left(a+b\right)\)
\(=a^3+a^2+b^3-b^2-a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a-b\right)\left(a+b\right)-a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2+a-b-a^2b^2\right)\)
b: \(=a^m\cdot a^3+2\cdot a^m\cdot a^2+a^m\)
\(=a^m\left(a^3+2a^2+1\right)\)
Phân tích đa thức thành nhân tử
a) (1-2x)(1+2x)-x(x+2)(x-2)
\(=1-4x^2-x\left(x^2-4\right)\)
\(=1-4x^2-x^3+4x\)
\(=\left(1-x^3\right)+\left(4x-4x^2\right)\)
\(=\left(1-x\right)\left(1+x+x^2\right)+4x\left(1-x\right)\)
\(=\left(1-x\right)\left(1+x+x^2+4x\right)\)
\(=\left(1-x\right)\left(x^2+5x+1\right)\)
\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8b^3a-8a^3b-12a^2b^2+6ab^3-b^4\)
\(=a^4+6a^3b+8b^3a-8a^3b-6ab^3-b^4\)
\(=\left(a^4-b^4\right)+\left(6a^3b-6ab^3\right)+\left(8b^3a-8a^3b\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a^2-b^2\right)+8ab\left(b^2-a^2\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a-b\right)\left(a+b\right)-8ab\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3+6a^2b+6ab^2-8a^2b-8ab^2\right)\)
\(=\left(a-b\right)\left(a^3-a^2b-ab^2+b^3\right)\)
\(=\left(a-b\right)\left[a^2\left(a-b\right)-b^2\left(a-b\right)\right]\)
\(=\left(a-b\right)^3\left(a+b\right)\)
a(b2-c2) - b(a2-c2) + c(a2-b2)
= a(b2-c2) - a2b + bc2 + a2c - b2c
= a(b+c)(b-c) + a2(c-b) + bc(c-b)
= a(b+c)(b-c) - a2(b-c) - bc(b-c)
= (b-c)[ a(b+c) - a2 - bc]
= (b-c)[ ab +ac - a2 - bc]
= (b-c)[ a(b-a) + c(a-b) ]
= (b-c)[ c(a-b) - a(a-b) ]
= (b-c)(a-b)(c-a)
= (a-b)(b-c)(c-a)
Bài 2:
a) =a2b - a2c + b2c - ab2 + ac2 - bc2
=(a2b - bc2) - (a2c - ac2) + (b2c - ab2)
=b(a-c)(a+c) - ac(a-c) - b2(a-c)
=(a - c)(ab -bc - ac - b2)
b)=(1 - 2a + a2) - (b2 - 2bc + c2)
=(1 - a)2 - (b - c)2
=(c - b - a + 1)(b - c - a + 1)