a) \(x-\dfrac{15}{x}=2\) (ĐKXĐ: \(x\ne0\))
\(\Leftrightarrow x^2-15=2x\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\) (TMĐK)
Vậy: \(S=\left\{5;-3\right\}\)
b) \(\dfrac{1}{x+1}-\dfrac{1}{x-1}=1\) (ĐKXĐ: \(x\ne\pm1\))
\(\Leftrightarrow\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow x-1-x-1-x^2+1=0\)
\(\Leftrightarrow x^2=-1\) (vô lí)
Vậy phương trình vô nghiệm.
d) \(x^5-x^3-x^2+1=0\)
\(\Leftrightarrow x^2\left(x^3-1\right)-\left(x^3-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2+x+1\right)\left(x+1\right)=0\)
Vì \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{1;-1\right\}\)
=>\(x^2-15=2x=>x^2-2x-15=0=>\left(x-5\right)\left(x+3\right)=>\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
