Phân tích đa thức thành nhân tử:
a^3+b^3+c^3-3abcb:(a+b+c)^3-a^3-b^3-c^3
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\(=\left(a+b\right)^3-3ab\left(a+b\right)-c^3+3abc\)
\(=\left(a+b\right)^3-c^3-3ab\left(a+b-c\right)\)
\(=\left(a+b-c\right)\left[\left(a+b\right)^2+c\left(a+b\right)+c^2\right]-3ab\left(a+b-c\right)\)
\(=\left(a+b-c\right)\left(a^2+b^2+c^2-ab+ac+bc\right)\)
`a, 8x^3 - 1 = (2x-1)(4x^2 + 2x - 1)`
`b, x^3 + 27y^3 = (x+3y)(x^3 - 3xy + 9y^2)`
`c, x^3 - y^6 = (x-y^2)(x+xy^2 + y^4)`
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)
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) \(x^3-4x^2+3x=x\left(x^2-4x+3\right)=x\left(x-1\right)\left(x-3\right)\)
\(A=4x\left(x^2-2x+1\right)=4x\left(x-1\right)^2\\ B=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\\ C=\left(x-2\right)\left(x^2+2x+4\right)+3\left(x-2\right)=\left(x-2\right)\left(x^2+2x+7\right)\)
a) \(A=4x\left(x^2-2x+1\right)=4x\left(x-1\right)^2\)
b) \(B=\left(x^2-2xy+y^2\right)-16=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\)
c) \(C=\left(x-2\right)\left(x^2+2x+4\right)+3\left(x-2\right)=\left(x-2\right)\left(x^2+2x+7\right)\)
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)
`a, P = 2x(3 - x^2)`
`b, Q = 5x^2(x-3y)`
`c, R = xy(3x^2y^2 - 6y^2z + 1)`
a) \(P=6x-2x^3\)
\(P=2x\left(3+x^2\right)\)
b) \(Q=5x^3-15x^2y\)
\(Q=5x^2\left(x-3y\right)\)
c) \(R=3x^3y^3-6xy^3z+xy\)
\(R=xy\left(3x^2y^2-6y^2z+1\right)\)
a)\(a^3+b^3+c^3-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
b)\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
a^3+b^3+c^3-3abc/(a+b+c)^a-a^3-b^3-c^3