Ta có : \(6x^4+7x^3-36x^2-7x+6=0\)
\(6x^4-12x^3+19x^3-38x^2+2x^2-4x-3x+6=0\)
\(6x^3\left(x-2\right)+19x^2\left(x-2\right)+2x\left(x-2\right)-3\left(x-2\right)=0\)
\(\left(x-2\right)\left(6x^3+19x^2+2x-3\right)=0\)
\(\left(x-2\right)\left(6x^3+18x^2+x^2+3x-x-3\right)=0\)
\(\left(x-2\right)\left(x+3\right)\left(6x^2+x-1\right)=0\)
\(\left(x-2\right)\left(x+3\right)\left(2x+1\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\\2x+1=0\\3x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\\x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Cách 2 : A = \(6x^4+7x^3-36x^2-7x+6=0\)
= \(6x^2\left(x^2+\dfrac{7}{6}x-6-\dfrac{7}{6x}+\dfrac{1}{x^2}\right)=0\)
= \(6x^2\left(\left(x^2+\dfrac{1}{x^2}-2\right)+\dfrac{7}{6}\left(x-\dfrac{1}{x}\right)-4\right)=0\)
= \(6x^2\left(\left(x-\dfrac{1}{x}\right)^2+\dfrac{7}{6}\left(x-\dfrac{1}{x}\right)-4\right)=0\)
Đặt \(A'=\left(x-\dfrac{1}{x}\right)^2+\dfrac{7}{6}\left(x-\dfrac{1}{x}\right)-4\)
và \(t=x-\dfrac{1}{x}\Rightarrow A'=t^2+\dfrac{7}{6}t-4\)
\(=t^2+2t.\dfrac{7}{12}+\left(\dfrac{7}{12}\right)^2-\left(\dfrac{7}{12}\right)^2-4\)
\(=\left(t+\dfrac{7}{12}\right)^2-\left(\dfrac{25}{12}\right)^2=\left(t-\dfrac{3}{2}\right)\left(t+\dfrac{8}{3}\right)\)
\(\Rightarrow A=6x^2\left(x-\dfrac{1}{x}-\dfrac{3}{2}\right)\left(x-\dfrac{1}{x}+\dfrac{8}{3}\right)=0\)
\(=6x^2\left(\dfrac{x^2-1-\dfrac{3}{2}x}{x}\right)\left(\dfrac{x^2-1+\dfrac{8}{3}x}{x}\right)=0\)
\(=\dfrac{6x^2\left(\left(x-\dfrac{3}{4}\right)^2-\left(\dfrac{5}{4}\right)^2\right)\left(\left(x+\dfrac{4}{3}\right)^2-\left(\dfrac{5}{3}\right)^2\right)}{x^2}=0\)
\(=6\left(x-2\right)\left(x+\dfrac{1}{2}\right)\left(x-\dfrac{1}{3}\right)\left(x+3\right)=0\)
Rồi giải ra các nghiệm trên nha !!! ( mình mỏi tay quá, phần sau các bạn làm như cách 1 nha )
.................. Làm sao nhỉ ??? Em ko thua đâu ! Đợi chút , em tìm cách giải !!!
Đây là phương trình có hệ số đối xứng bậc 4 ??? Đúng ko ???
Pt có 4 nghiệm :
X = 2 ; -3 ; -1/2 ; 1/3
