\(x^4+x^2+4x-3=0\)
\(\Leftrightarrow x^4+2x^2+1-x^2+4x-4=0\)
\(\Leftrightarrow\left(x^2+1\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x^2+1-x+2\right)\left(x^2+1+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+3\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow x^2+x-1=0\)
\(\Leftrightarrow x=\dfrac{-1\pm\sqrt{5}}{2}\)
Cách khác: (Tổng quát)
\(x^4+x^2+4x-3=0\)
\(\Leftrightarrow x^4+2x^2y+y^2-2x^2y-y^2+x^2+4x-3=0\)
\(\Leftrightarrow\left(x^2+y\right)^2-\left(2y-1\right)\left[x^2-\dfrac{4x}{2y-1}+\dfrac{4}{\left(2y-1\right)^2}\right]+\dfrac{4}{2y-1}-y^2-3=0\)
\(\Leftrightarrow\left(x^2+y\right)^2-\left(2y-1\right)\left(x-\dfrac{2}{2y-1}\right)^2+\dfrac{4}{2y-1}-y^2-3=0\left(1\right)\)
Ta mong muốn: \(\dfrac{4}{2y-1}-y^2-3=0\)
\(\Leftrightarrow2y^3-y^2-6y+7=0\)
\(\Leftrightarrow y=1\)
Khi đó:
\(\left(1\right)\Leftrightarrow\left(x^2+y\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x^2-x+3\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow x^2+x-1=0\)
\(\Leftrightarrow x=\dfrac{-1\pm\sqrt{5}}{2}\)
\(x^4+x^2+4x-3=0\)
\(\Leftrightarrow x^4+2x^2+1-x^2+4x-4=0\)
\(\Leftrightarrow\left(x^2+1\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x^2+1-x+2\right)\left(x^2+1+x-2\right)=0\)
\(\Leftrightarrow x^2+x-1=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-1\right)=5\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{5}}{2}\\x_2=\dfrac{-1+\sqrt{5}}{2}\end{matrix}\right.\)
