phân tích đa thức thành nhân tử:
a^3+b^3-c^3+3abc
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a3 + b3 + c3 - 3abc = (a + b)3 + c3 - 3abc - 3ab(a + b)
= (a + b + c)(a2 + b2 + 2ab - ac - bc + c2) - 3ab(a + b + c)
= (a + b + c)(a2 + b2 + c2 - ab - ac - bc)
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)^3-3c\left(a+b\right)\left(a+b+c\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b+c\right)^2-3ac-3bc-3ab\right]\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ac+2bc+2ab-3ac-3bc-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
Bài 1:
a: \(4a^2-6b=2\left(2a^2-3b\right)\)
b: \(m^3n-2m^2n^2-mn\)
\(=mn\left(m^2-2mn-1\right)\)
Bài 1:
a) \(4a^2-6b=2\left(a^2-3b\right)\)
b) \(=mn\left(m^2-2mn-1\right)\)
Bài 2:
a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)
b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)
a) \(x^2-9+2\left(x+3\right)=\left(x-3\right)\left(x+3\right)+2\left(x+3\right)=\left(x+3\right)\left(x-3+2\right)=\left(x+3\right)\left(x-1\right)\)
b) \(x^2-10x+25-3\left(x-5\right)=\left(x-5\right)^2-3\left(x-5\right)=\left(x-5\right)\left(x-5-3\right)=\left(x-5\right)\left(x-8\right)\)
c) \(x^3-4x^2+3x=x\left(x^2-4x+3\right)=x\left(x-1\right)\left(x-3\right)\)
Ta có
a3+b3+c3-3abc
=(a+b)3-3ab(a+b)+c3-3abc
=[(a+b)3+c3]-3ab(a+b+c)
=(a+b+c)[(a+b)2-c(a+b)+c2]-3ab(a+b+c)
=(a=b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)
=(a+b+c)(a2+2ab+b2-ac-bc+c2-3ab)
=(a+b+c)(a2+b2+c2-ab-ac-bc)
\(=\left(a+b\right)^3-3ab\left(a+b\right)-c^3+3abc\)
\(=\left(a+b\right)^3-c^3-3ab\left(a+b-c\right)\)
\(=\left(a+b-c\right)\left[\left(a+b\right)^2+c\left(a+b\right)+c^2\right]-3ab\left(a+b-c\right)\)
\(=\left(a+b-c\right)\left(a^2+b^2+c^2-ab+ac+bc\right)\)