a/ đkxđ: \(x^2-8x+15>0\)
\(\Leftrightarrow x^2-8x+16-1>0\)
\(\Leftrightarrow\left(x-4\right)^2>1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4>1\Rightarrow x>5\\x-4< -1\Leftrightarrow x< 3\end{matrix}\right.\)
Vậy x < 3 hoặc x > 5
b/ đkxđ: \(2-x^2\ge0\)\(\Leftrightarrow-\sqrt{2}\le x\le\sqrt{2}\)
vậy.........
c/ đkxđ: \(\dfrac{2x-1}{1-x}\ge0\) và 1 - x ≠ 0
=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\1-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\1-x< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x>1\end{matrix}\right.\end{matrix}\right.\)
=> \(\dfrac{1}{2}\le x< 1\)
Vậy.........
