\(6x^4-5x^3-38x^2-5x+6=0\)
\(\Leftrightarrow6x^4-12x^3+17x^3-34^2-4x^2+8x-3x+6=0\)
\(\Leftrightarrow6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x^3+18x^2-4x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x^3+18x^2-x^2-3x-x-3=0\right)\)
\(\Leftrightarrow\left(x-2\right)\left[6x^2\left(x+3\right)-x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(6x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(6x^2-3x+2x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left[6x\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{2}\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-\frac{1}{2}\right)\left(6x+2\right)=0\)
Phương trình đã cho tương đương với: \(\left(6x^4+6\right)-\left(5x^3+5x\right)-38x^2=0\)
Dễ thấy \(x=0\)không phải là nghiệm của phương trình, chia cả 2 vế cho \(x^2\)ta được: \(6\left(x^2+\frac{1}{x^2}\right)-5\left(x+\frac{1}{x}\right)-38=0\)Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
\(\Rightarrow6\left(t^2-2\right)-5t-38=0\Leftrightarrow6t^2-5t-50=0\Leftrightarrow\hept{\begin{cases}t=\frac{10}{3}\\t=-\frac{5}{2}\end{cases}}\)
-Với \(t=\frac{10}{3}\Rightarrow x+\frac{1}{x}=\frac{10}{3}\Leftrightarrow3x^2-10x+3\Leftrightarrow\hept{\begin{cases}x=3\\x=\frac{1}{3}\end{cases}}\)
-Với \(t=-\frac{5}{2}\Rightarrow x+\frac{1}{x}=-\frac{5}{2}\Leftrightarrow2x^2+5x+2\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=-2\end{cases}}\)
Vậy phương trình có 4 nghiệm: \(x=\left\{-2;-\frac{1}{2};\frac{1}{3};3\right\}\)
