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\(\left(4x-1\right)\sqrt{x^3+1}-1=2x^3+2x\)
Đặt \(\sqrt{x^3+1}=t\ge0\)
\(\Leftrightarrow\left(4x-1\right)t-1=2t^2-2+2x\)
\(\Leftrightarrow2t^2-\left(4x-1\right)t+2x-1=0\)
\(\Delta=\left(4x-1\right)^2-8\left(2x-1\right)=16x^2-24x+9=\left(4x-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{4x-1+4x-3}{4}=2x-1\\t=\frac{4x-1-4x+3}{4}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^3+1}=2x-1\left(x\ge\frac{1}{2}\right)\\\sqrt{x^3+1}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^3-4x^2+4x=0\\x^3=-\frac{3}{4}\end{matrix}\right.\) \(\Leftrightarrow...\)
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