Question

Phân tích thành nhân tử:

\(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-2abc-a^3-b^3-c^3\)

\(x^4-7x^3+14x^2-7x+1\)

\(x^{12}+x^6+1\)

 

Answer 1

c)x¹²+x⁶+1

Lần lượt thêm và bớt x⁹; x³;x⁶ ta đc:

=x¹²+x⁹-x⁶-x⁹-x⁶-x³+x⁶+x³+1

=x⁶(x⁶+x³+1)-x³(x⁶+x³+1)+(x⁶+x³+1)

=(x⁶-x³+1)(x⁶+x³+1)

Answer 2

b)x⁴-7x³-14x²-7x+1

=x⁴-3x³+x²-4x³+12x²-4x+x²-3x+1

=x²(x²-3x+1)-4x(x²-3x+1)+(x²-3x+1)

=(x²-4x+1)(x²-3x+1)

 

Answer 3

\(A=x^4-7x^3+14x^2-7x+1\)

Giả sử:

\(A=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)

\(=x^4+cx^3+dx^2+ax^3+acx^2+adx+bx^2+bcx+bd\)

\(=x^4+\left(a+c\right)x^3+\left(d+ac+b\right)x^2+\left(ad+bc\right)x+bd\)

Ta có:

\(\begin{cases}a+c=-7\\d+ac+b=14\\ad+bc=-7\\bd=1\end{cases}\)\(\Rightarrow\begin{cases}a=-4\\b=1\\c=-3\\d=1\end{cases}\)

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