Question

Tìm TSĐ:

y=\(\dfrac{x-5}{\left|x^2-x-2\right|+\left|x+1\right|}\)

y= \(\sqrt{\dfrac{x}{1-x}}+\sqrt{2x-1}\)

y= \(\dfrac{x^2-4\sqrt{5-2x}}{3-x\left(x+2\right)}\)

y= \(\sqrt{x+3-2\sqrt{x+2}}\)

Answer 1

a: ĐKXĐ: (x-2)(x+1)<>0 và x+1<>0

=>\(x< >-1\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}\dfrac{x}{1-x}>=0\\2x-1>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0< =x< 1\\x>=\dfrac{1}{2}\end{matrix}\right.\)

c: ĐKXĐ: 5-2x>=0 và 3-x(x+2)<>0

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{5}{2}\\x^2+2x-3< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{5}{2}\\x\notin\left\{-3;1\right\}\end{matrix}\right.\)