Lời giải:
\((x^2-4)^2=8x+1\)
\(\Leftrightarrow (x^2-4)^2-9=8x-8\)
\(\Leftrightarrow (x^2-4)^2-3^2=8(x-1)\)
\(\Leftrightarrow (x^2-7)(x^2-1)=8(x-1)\)
\(\Leftrightarrow (x^2-7)(x-1)(x+1)-8(x-1)=0\)
\(\Leftrightarrow (x-1)(x^3+x^2-7x-7-8)=0\)
\(\Leftrightarrow (x-1)(x^3+x^2-7x-15)=0\)
\(\Leftrightarrow (x-1)[x^2(x-3)+4x(x-3)+5(x-3)]=0\)
\(\Leftrightarrow (x-1)(x-3)(x^2+4x+5)=0\)
\(\Rightarrow \left[\begin{matrix} x-1=0\\ x-3=0\\ x^2+4x+5=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=1\\ x=3\\ (x+2)^2+1=0(\text{vô lý})\end{matrix}\right.\)
Vậy PT có nghiệm \(x=1\) hoặc $x=3$
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