Question

Tìm điều kiện để các biểu thức sau xác định

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

b)\(2.\sqrt{1-2x}-\dfrac{\sqrt{3}-1}{4}\)

c)\(\sqrt{x+1}-\sqrt{x-2}\)

d)\(\sqrt{2-3x}-\sqrt{1-2x}\)  

e)\(2.\sqrt{\sqrt{3}-2x}+\dfrac{1}{x-1}\)

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

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

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

 

Answer 1

a, \(x+1\ge0\Leftrightarrow x\ge-1\)

b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)

c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)

Answer 2

d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)

e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)

Answer 3

f, \(\left\{{}\begin{matrix}x-\dfrac{\sqrt{3}}{2}\ge0\\x\ge0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ge0\\x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\)

g, \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+2\ne0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Answer 4

a: ĐKXĐ: \(x\ge-1\)

b: ĐKXĐ: \(x\le\dfrac{1}{2}\)

c: ĐKXĐ: \(x\ge2\)

d: ĐKXĐ: \(x\le\dfrac{1}{2}\)

Answer 5

h, \(x\ge0\)