a: \(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{x}{x^2-2x+1}-\dfrac{1}{x^2-1}\right)\)

\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)

\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x\left(x+1\right)-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)

\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{x^2+x-x+1}{x-1}\)

\(=\dfrac{1-x}{x-1}=-1\)

b: \(\dfrac{x}{6-x}+\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right):\dfrac{2x-6}{x^2+6x}\)

\(=\dfrac{x}{6-x}+\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}\)

\(=\dfrac{x}{6-x}+\dfrac{x^2-x^2+12x-36}{x-6}\cdot\dfrac{1}{2\left(x-3\right)}\)

\(=\dfrac{x}{6-x}+\dfrac{12\left(x-3\right)}{2\left(x-3\right)\left(x-6\right)}\)

\(=\dfrac{x}{6-x}+\dfrac{6}{x-6}=\dfrac{-x+6}{x-6}=-1\)

\(\left\{{}\begin{matrix}16-x^2\ge0\\2x+1>0\\x^2-8x+14\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4\le x\le4\\x>-\dfrac{1}{2}\\\left[{}\begin{matrix}x\ge4+\sqrt{2}\\x\le4-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-\dfrac{1}{2}< x\le4-\sqrt{2}\)

xác định \(< =>\left\{{}\begin{matrix}\sqrt{16-x^2}\ge0\\\sqrt{2x+1}>0\\\sqrt{x^2-8x+14}\ge0\end{matrix}\right.\)

\(< =>\left\{{}\begin{matrix}-4\le x\le4\\x>-\dfrac{1}{2}\\\left[{}\begin{matrix}x\le4-\sqrt{2}\\x\ge4_{ }+\sqrt{2}\end{matrix}\right.\\\end{matrix}\right.\)\(< =>-\dfrac{1}{2}< x\le4-\sqrt{2}\)

ĐKXĐ: \(\left\{{}\begin{matrix}16-x^2\ge0\\2x+1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2\le16\\x>-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le-4\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge4\\x>-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x\ge4\)

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10