Ta có : \(x^4-4x^3+6x^2-4x-15=0\)
=> \(x^4-3x^3-x^3+3x^2+3x^2-9x+5x-15=0\)
=> \(x^3\left(x-3\right)-x^2\left(x-3\right)+3x\left(x-3\right)+5\left(x-3\right)=0\)
=> \(\left(x-3\right)\left(x^3-x^2+3x+5\right)=0\)
=> \(\left(x-3\right)\left(x^3+x^2-2x^2-2x+5x+5\right)=0\)
=> \(\left(x-3\right)\left(x^2\left(x+1\right)-2x\left(x+1\right)+5\left(x+1\right)\right)=0\)
=> \(\left(x-3\right)\left(x+1\right)\left(x^2-2x+5\right)=0\)
=> \(\left(x-3\right)\left(x+1\right)\left(x^2-2x+1+4\right)=0\)
=> \(\left(x-3\right)\left(x+1\right)\left(\left(x-1\right)^2+4\right)=0\)
Mà \(\left(x-1\right)^2+4>0\)
=> \(\left(x-3\right)\left(x+1\right)=0\)
=> \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{3;-1\right\}\)
Phương trình tương đương:
\(\begin{array}{l} {x^4} - 4{x^3} + 6{x^2} - 4x + 1 = 16\\ \Leftrightarrow {\left( {x - 1} \right)^4} = 16\\ \Leftrightarrow {\left[ {{{\left( {x - 1} \right)}^2}} \right]^2} - \left( {{2^2}} \right) = 0\\ \Leftrightarrow \left[ {{{\left( {x - 1} \right)}^2} - {2^2}} \right]\left[ {{{\left( {x - 1} \right)}^2} + {2^2}} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} {\left( {x - 1} \right)^2} = 4 \Leftrightarrow \left[ \begin{array}{l} x - 1 = 2\\ x - 1 = - 2 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 3\\ x = - 1 \end{array} \right.\\ {\left( {x - 1} \right)^2} = - 4 (VN) \end{array} \right. \end{array}\)
\(\Leftrightarrow y\left(y+1\right)=x\left(x+1\right)\left(x^2+1\right)\)
