Question

1. với giá trị nào của x thì biểu thuéc sau có nghĩa

\(\sqrt{x^2-3}\)

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

\(\sqrt{\dfrac{1}{3-2xx}}\)

 

Answer 1

a, ĐKXĐ: \(x^2-3\ge0\Rightarrow x^2\ge3\Rightarrow x\ge\sqrt{3}\)

b, \(\left\{{}\begin{matrix}x-2\ne0\\x-2\ge0\end{matrix}\right.\Rightarrow x-2>0\Rightarrow x>2\)

c, \(\left\{{}\begin{matrix}3-2x\ne0\\\dfrac{1}{3-2x}\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2x\ne3\\3-2x>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{3}{2}\\x< \dfrac{3}{2}\end{matrix}\right.\)

Answer 2

\(\sqrt{x^2-3}\)

ĐKXĐ: x > 1

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

ĐKXĐ: x > 2

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

ĐKXĐ: x < 1,224744871 \(\approx\) 1,22

Answer 3

a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge\sqrt{3}\\x\le-\sqrt{3}\end{matrix}\right.\)

b: ĐKXĐ: x>2

c: ĐKXĐ: \(x< \dfrac{3}{2}\)