Question

[Nguyễn tuấn nghĩa]

Giải phương trình \(x^4-4x^3+3x^2+2x-6=0\)

Answer 1

[Thiên Đạo Pain]

guitykamikk kém quá :) ko giải dc pt bậc 3 à :)) hhahah , nhìn chúa làm đây này 

\(x^4-4x^3+4x^2=4x^2-3x^2-2x+6\)

\(\left(x^2-2x\right)^2=x^2-2x+6\)

\(\left(x^2-2x\right)^2+2\left(x^2-2x\right)m+m^2=x^2-2x+6+2\left(x^2-2x\right)m+m^2\)

\(\left(x^2-2x+m\right)^2=x^2-2x+6+2x^2m-4xm+m^2\)

\(\left(x^2-2x+m\right)^2=x^2\left(1+2m\right)-2x\left(1+2m\right)+\left(6+m^2\right)\)

\(\Delta'=\left(1+2m\right)^2-\left(6+m^2\right)\left(1+2m\right)\)

\(\Delta`=1+4m+4m^2-\left(6+12m+m^2+2m^3\right)\)

\(\Delta`=1+4m+4m^2-6-12m-m^2-2m^3\)

\(\Delta=-2m^3+3m^2-8m-5\)

\(\Delta=-m^2\left(2m+1\right)+2m\left(2m+1\right)-5\left(2m+1\right)\)

\(\Delta=\left(2m+1\right)\left(-m^2+2m-5\right)\)

\(m=-\frac{1}{2}\)

thay m=-1/2 vào ta được

\(\left(x^2-2x-\frac{1}{2}\right)^2=6+\frac{1}{4}\)

\(\left(x^2-2x-\frac{1}{2}\right)^2=\frac{25}{4}\)

\(\hept{\begin{cases}x^2-2x-\frac{1}{2}=\frac{5}{2}\\x^2-2x-\frac{1}{2}=-\frac{5}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2-2x-3=0\\x^2-2x+2>0\end{cases}}\)

\(x^2-2x-3=0\)

\(\Delta`=1+3=4\)

\(\hept{\begin{cases}x_1=1+\sqrt{4}=3\\1-\sqrt{4}=-1\end{cases}}\)

vậy nghiệm của pt là 3 , -1  

Answer 2

[_Guiltykamikk_]

\(x^4-4x^3+3x^2+2x-6=0\)

\(\Leftrightarrow\left(x^4-3x^3\right)-\left(x^3-3x^2\right)+\left(2x-6\right)=0\)

\(\Leftrightarrow x^3\left(x-3\right)-x^2\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x^3-x^2+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^3-x^2+2=0\\x-3=0\end{cases}}\)