a)
Đặt \(\sqrt[3]{x}=a\). Khi đó pt trở thành:
\(a^2-3a=20\)
\(\Leftrightarrow a^2-3a+\left(\frac{3}{2}\right)^2=\frac{89}{4}\)
\(\Leftrightarrow (a-\frac{3}{2})^2=\frac{89}{4}\)
\(\Rightarrow \left[\begin{matrix} a-\frac{3}{2}=\frac{\sqrt{89}}{2}\\ a-\frac{3}{2}=\frac{-\sqrt{89}}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} a=\frac{3}{2}+\frac{\sqrt{89}}{2}\\ a=\frac{3}{2}-\frac{\sqrt{89}}{2}\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=a^3=\left(\frac{3+\sqrt{89}}{2}\right)^3\\ x=a^3=\left(\frac{3-\sqrt{89}}{2}\right)^3\end{matrix}\right.\)
b)
Đặt \(\left\{\begin{matrix} \sqrt{x^2+1}=a\\ 2x-1=b\end{matrix}\right.(a>0)\)
Khi đó, pt trở thành:
\(a(2b+1)=2a^2+b\)
\(\Leftrightarrow (2a^2-2ab)-(a-b)=0\)
\(\Leftrightarrow 2a(a-b)-(a-b)=0\)
\(\Leftrightarrow (2a-1)(a-b)=0\)
Từ đây xét các TH:
TH1: \(2a-1=0\Leftrightarrow a=\frac{1}{2}\Leftrightarrow \sqrt{x^2+1}=\frac{1}{2}\)
\(\Leftrightarrow x^2=\frac{1}{4}-1=\frac{-3}{4}< 0\) (vô lý)
TH2: \(a-b=0\Leftrightarrow \sqrt{x^2+1}=2x-1\)
\(\Rightarrow \left\{\begin{matrix} x^2+1=(2x-1)^2\\ 2x-1\geq 0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} 3x^2-4x=0\\ x\geq \frac{1}{2}\end{matrix}\right.\Rightarrow x=\frac{4}{3}\)
Vậy.......
