ta có : \(x^4-10x^3-15x^2+20x+4=0\) (*)
\(\Leftrightarrow x^4-x^3-9x^3+9x^2-24x^2+24x-4x+4=0\)
\(\Leftrightarrow x^3\left(x-1\right)-9x^2\left(x-1\right)-24x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^3-9x^2-24x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^3-11x^2-2x+2x^2-22x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x\left(x^2-11x-2\right)+2\left(x^2-11x-2\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-11x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x^2-11x-2=0\left(xétsau\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
ta có : \(x^2-11x-2=0\) (1)
\(\Delta=11^2-4.1.\left(-2\right)=121+8=129>0\)
\(\Rightarrow\) phương trình (1) có 2 nghiệm phân biệt
\(x_1=\dfrac{11+\sqrt{129}}{2}\) ; \(x_2=\dfrac{11-\sqrt{129}}{2}\)
vậy phương trình (*) có 4 nghiệm phân biệt \(x=1;x=-2;x=\dfrac{11+\sqrt{129}}{2};x=\dfrac{11-\sqrt{129}}{2}\)
