Question

x³=2\(\sqrt[3]{2x-1}\)-1

 

Answer 1

Đặt \(\sqrt[3]{2x-1}=a\).

Ta có hpt:

\(\left\{{}\begin{matrix}x^3=2a-1\\a^3=2x-1\end{matrix}\right.\).

Nếu x > a thì \(x^3>a^3\Rightarrow2a-1>2x-1\Rightarrow a>x\) (vô lí).

Nếu x < a thì \(x^3< a^3\Rightarrow2a-1< 2x-1\Rightarrow a< x\) (vô lí).

Do đó a = x.

 Khi đó \(x^3=2x-1\Leftrightarrow x^3-2x+1=0\Leftrightarrow\left(x-1\right)\left(x^2+x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{\pm\sqrt{5}-1}{2}\end{matrix}\right.\).

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