a/ - Với \(x\ge1\):
\(\Leftrightarrow x^2-3x+2+x-1=0\)
\(\Leftrightarrow x^2-2x+1=0\Rightarrow x=1\)
- Với \(x< 1\)
\(\Leftrightarrow x^2-3x+2+1-x=0\)
\(\Leftrightarrow x^2-4x+3=0\Rightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=3\left(l\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=1\)
b/ ĐKXĐ: ...
\(\Leftrightarrow8\left(x^2+\frac{1}{x^2}+2\right)+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}+2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x^2+\frac{1}{x^2}\right)+16+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow\left(x+4\right)^2=16\Rightarrow\left[{}\begin{matrix}x+4=4\\x+4=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-8\end{matrix}\right.\)
