Lời giải:
Ta có: \(x^3-5x+4=0\)
\(\Leftrightarrow (x^3-x)-(4x-4)=0\)
\(\Leftrightarrow x(x^2-1)-4(x-1)=0\)
\(\Leftrightarrow (x-1)[x(x+1)-4]=0\)
\(\Leftrightarrow \left[\begin{matrix} x-1=0\\ x^2+x-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ (x+\frac{1}{2})^2=\frac{17}{4}\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=1\\ x=\pm \frac{\sqrt{17}}{2}-\frac{1}{2}\end{matrix}\right.\)
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