a, ĐKXĐ:\(\left\{{}\begin{matrix}x+4\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-4\\x\ne1\end{matrix}\right.\)
b, ĐKXĐ:\(\left\{{}\begin{matrix}5\left(x-2\right)\ne0\\3\left(x+2\right)\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\\x\ne\pm2\end{matrix}\right.\Leftrightarrow x\ne\pm2\)
c, ĐKXĐ:\(\left\{{}\begin{matrix}4x^2-8x+7\ne0\\4x^2-10x+7\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4\left(x^2-2x+\dfrac{7}{4}\right)\ne0\\4\left(x^2-\dfrac{10}{4}x+\dfrac{7}{4}\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-2x+1\right)+\dfrac{3}{4}\ne0\\\left(x^2-2.\dfrac{10}{8}.x+\dfrac{25}{16}\right)+\dfrac{3}{16}\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2+\dfrac{3}{4}\ne0\left(luôn.đúng\right)\\\left(x-\dfrac{5}{4}\right)^2+\dfrac{3}{16}\ne0\left(luôn.đúng\right)\end{matrix}\right.\)
\(\Leftrightarrow x\in R\)
a, đkxđ \(\left\{{}\begin{matrix}x+4\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-4\\x\ne1\end{matrix}\right.\)
b, đkxđ \(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
c, đkxđ \(\left\{{}\begin{matrix}4x^2-8x+7\ne0\\4x^2-10x+7\ne0\end{matrix}\right.\)*luôn đúng *
Vậy \(x\in R\)
