a/ ĐK: \(x\ge1\)
\(2\sqrt{x-1}=x^2-2x+1-\left(x-1\right)+2\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)-2\sqrt{x-1}+2=0\)
Đặt \(\sqrt{x-1}=t\ge0\) phương trình trở thành:
\(t^4-t^2-2t+2=0\Leftrightarrow t^2\left(t^2-1\right)-2\left(t-2\right)=0\)
\(\Leftrightarrow\left(t^3+t\right)\left(t-1\right)-2\left(t-1\right)=0\Leftrightarrow\left(t-1\right)\left(t^3+t-2\right)=0\)
\(\Leftrightarrow\left(t^2+2t+2\right)\left(t-1\right)^2=0\Rightarrow t-1=0\Rightarrow t=1\) (do \(t^2+2t+2>0\) \(\forall t\))
\(\Rightarrow\sqrt{x-1}=1\Rightarrow x=2\)
b/ ĐK: \(x\ge-\dfrac{1}{2}\)
Ta có \(\sqrt{8x^3+1}=\sqrt{\left(2x+1\right)\left(4x^2-2x+1\right)}\)'
Đặt \(\sqrt{2x+1}=a\ge0\) ; \(\sqrt{4x^2-2x+1}=b>0\) , phương trình trở thành:
\(a+3b=3+ab\Leftrightarrow a-3+3b-ab=0\)
\(\Leftrightarrow\left(a-3\right)-b\left(a-3\right)=0\Leftrightarrow\left(a-3\right)\left(1-b\right)=0\Rightarrow\left[{}\begin{matrix}a-3=0\\1-b=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{4x^2-2x+1}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x+1=9\\4x^2-2x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=4\\x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
