Question

Answer 1

2, ĐKXĐ: \(1-3x\ge0\\ \Rightarrow x\ge\dfrac{1}{3}\)

\(\sqrt{1-3x}-2< 0\\ \Leftrightarrow\sqrt{1-3x}< 2\\ \Leftrightarrow1-3x< 4\\ \Leftrightarrow3x< -3\\ \Leftrightarrow x< -1\)

\(\Leftrightarrow-\dfrac{1}{3}\le x< -1\)

Answer 2

\(\sqrt{6}=\sqrt{5+1}\)

Answer 3

Bài 1:

a.

\(\sqrt{5}+1>\sqrt{4}+1=3=\sqrt{9}>\sqrt{6}\)

b.

\(\sqrt{3}-2=\frac{-1}{\sqrt{3}+2}\)

\(2-\sqrt{5}=\frac{1}{2+\sqrt{5}}\)

Mà \(\frac{-1}{\sqrt{3}+2}< \frac{-1}{2+\sqrt{5}}\Rightarrow \sqrt{3}-2< 2-\sqrt{5}\)

Answer 4

Bài 2: 

BPT \(\Leftrightarrow \left\{\begin{matrix} 1-3x\geq 0\\ \sqrt{1-3x}< 2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq \frac{1}{3}\\ 1-3x< 4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq \frac{1}{3}\\ x> -1\end{matrix}\right.\)

\(\Leftrightarrow \frac{1}{3}\geq x> -1\)