Question

[Trần Duy]

1.Biết x³-x=6,tính: A=x⁶-2x⁴+x³+x²-x

2.Phân tích đa thức thành nhân tử

a)a(b² +c²+bc)+b(c²+a²+ac)+c(a²+b²+ab)

b)(a+b+c)(ab+bc+ca)-abc

Answer 1

[KZ]

1. \(A=x^6-x^4+x^3-x-x^4+x^2=x^3\left(x^3-x\right)+\left(x^3-x\right)-x\left(x^3-x\right)=\left(x^3-x\right)^2+\left(x^3-x\right)\)

Thay \(x^3-x=6\) vào A, ta được:

\(A=36+6=42\)

KL : A=42

2.

a) đa thức đã cho \(=ab^2+ac^2+abc+bc^2+ba^2+abc+ca^2+cb^2+abc\)

\(=\left(ab^2+ba^2+abc\right)+\left(ac^2+ca^2+abc\right)+\left(bc^2+cb^2+abc\right)\)

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

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

b) đa thức đã cho \(=\left(a^2c+abc\right)+\left(abc+b^2c\right)+\left(ac^2+bc^2\right)+\left(a^2b+ab^2\right)\)

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

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

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

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

\(=\left(a+c\right)\left(b+c\right)\left(a+b\right)\)