Question

Tìm ĐKXĐ:

A = \(\sqrt{2-4x}\)

B = \(\sqrt{\dfrac{-3}{x-1}}\)+ \(\sqrt{x^2+4}\)

C = \(\dfrac{2x+3}{\sqrt{4x^2-12x+9}}\)

Answer 1

a, ĐKXĐ: \(2-4x\ge0\)

\(\Rightarrow x\le\dfrac{1}{2}\)

b, ĐKXĐ: \(\left\{{}\begin{matrix}\dfrac{-3}{x-1}>0\\x^2+4\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1< 0\\x\in R\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 1\\x\in R\end{matrix}\right.\)

(Do ta có: \(x^2+4\ge0\) \(\left(\forall x\in R\right)\))

c, ĐKXĐ: \(4x^2-12x+9>0\) (do biểu thức căn dưới mẫu)

\(\Rightarrow\left(2x-3\right)^2>0\)

\(\Rightarrow x\ne\dfrac{3}{2}\)