Question

Giải 2 phương trình sau(pt hồi quy)

\[ x^4 - 3x^3 - 6x^2 + 3x + 1 = 0 \]

\[ 2x^4 - 21x^3 + 74x^2 - 105x + 50 = 0 \]

Answer 1

a: \(x^4-3x^3-6x^2+3x+1=0\)

=>\(x^4+x^3-x^2-4x^3-4x^2+4x-x^2-x+1=0\)

=>\(\left(x^2+x-1\right)\left(x^2-4x-1\right)=0\)

=>\(\left[{}\begin{matrix}x^2+x-1=0\\x^2-4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1\pm\sqrt{5}}{2}\\x=2\pm\sqrt{5}\end{matrix}\right.\)

b: \(2x^4-21x^3+74x^2-105x+50=0\)

=>\(2x^4-2x^3-19x^3+19x^3+55x^2-55x-50x+50=0\)

=>\(\left(x-1\right)\cdot\left(2x^3-19x^2+55x-50\right)=0\)

=>\(\left(x-1\right)\left(2x^3-4x^2-15x^2+30x+25x-50\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)\left(2x^2-15x+25\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)\left(2x^2-10x-5x+25\right)=0\)

=>(x-1)(x-2)(x-5)(2x-5)=0

=>\(\left[{}\begin{matrix}x-1=0\\x-2=0\\x-5=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=5\\x=\dfrac{5}{2}\end{matrix}\right.\)