a/ \(\Leftrightarrow2\left(x^2+1\right)-\left(4x-1\right)\sqrt{x^2+1}+2x-1=0\)
Đặt \(\sqrt{x^2+1}=a\ge1\)
\(\Rightarrow2a^2-\left(4x-1\right)a+2x-1=0\)
\(\Delta=\left(4x-1\right)^2-8\left(2x-1\right)=\left(4x-3\right)^2\)
Phương trình có 2 nghiệm: \(\left[{}\begin{matrix}t=\frac{4x-1-4x+3}{4}=\frac{1}{2}< 1\left(l\right)\\t=\frac{4x-1+4x-3}{4}=2x-1\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2+1}=2x-1\) \(\left(x\ge\frac{1}{2}\right)\)
\(\Leftrightarrow x^2+1=4x^2-4x+1\)
\(\Leftrightarrow3x^2-4x=0\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=\frac{4}{3}\end{matrix}\right.\)
b/
Đặt \(\sqrt[3]{2x-1}=a\Rightarrow a^3+1=2x\)
Ta được hệ:
\(\left\{{}\begin{matrix}x^3+1=2a\\a^3+1=2x\end{matrix}\right.\)
\(\Rightarrow x^3-a^3=2a-2x\)
\(\Leftrightarrow\left(x-a\right)\left(x^2+ax+a^2\right)+2\left(x-a\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left(x^2+ax+a^2+2\right)=0\)
\(\Leftrightarrow\left(x-a\right)\left[\left(x+\frac{a}{2}\right)^2+\frac{3a^2}{4}+2\right]=0\)
\(\Leftrightarrow x-a=0\)
\(\Rightarrow x=\sqrt[3]{2x-1}\Leftrightarrow x^3-2x+1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x-1\right)=0\)
\(\Leftrightarrow...\)
