Question

1. Tìm các giá trị của x để các biểu thức sau có nghĩa :

A= \(\dfrac{1}{\sqrt{x^2+4x-5}}\)

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

C= \(\dfrac{1}{1-\sqrt{x^2-3}}\)

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

E=\(\sqrt{3x-1}-\sqrt{5x-3}+\sqrt{x^2+x+1}\)

Answer 1

a/ \(x^2+4x-5>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -5\end{matrix}\right.\)

b/ \(\left\{{}\begin{matrix}2x-1\ge0\\x-\sqrt{2x-1}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\\left\{{}\begin{matrix}x>0\\x^2>2x-1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\end{matrix}\right.\)

c/ \(\left\{{}\begin{matrix}x^2-3\ge0\\1-\sqrt{x^2-3}\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\sqrt{3}\\x\le-\sqrt{3}\end{matrix}\right.\\x\ne\pm2\end{matrix}\right.\)

d/ \(\left\{{}\begin{matrix}x+\dfrac{1}{x}\ge0\\-2x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>0\\x\le0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn

e/ \(\left\{{}\begin{matrix}3x-1\ge0\\5x-3\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\ge\dfrac{3}{5}\end{matrix}\right.\) \(\Rightarrow x\ge\dfrac{3}{5}\)