Question

Tìm giá trị của x:

a) \(\sqrt{2x}< \dfrac{1}{3}\)

b) \(\sqrt{-3x+\dfrac{1}{2}}\ge5\)

c) \(\sqrt{-2x+1}>7\)

d) \(\sqrt{2x-1}\le\dfrac{3}{2}\)

Answer 1

a.ĐKXĐ: \(x\ge0\)
 \(\sqrt{2x}< \dfrac{1}{3}\) \(\Leftrightarrow2x< \dfrac{1}{3}\Leftrightarrow6x< 1\Leftrightarrow x< \dfrac{1}{6}\)

b. ĐKXĐ: \(x\ge\dfrac{1}{6}\)

\(\sqrt{-3x+\dfrac{1}{2}}\ge5\Leftrightarrow-3x+\dfrac{1}{2}\ge25\Leftrightarrow x=-\dfrac{49}{6}\) 

c. ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{-2x+1}>7\) \(\Leftrightarrow-2x+1>49\Leftrightarrow x=-24\)

d. ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-1}\le\dfrac{3}{2}\Leftrightarrow2x-1\le\dfrac{9}{4}\Leftrightarrow x=\dfrac{13}{8}\)

 

Answer 2

a: Ta có: \(\sqrt{2x}< \dfrac{1}{3}\)

\(\Leftrightarrow2x< \dfrac{1}{9}\)

\(\Leftrightarrow x< \dfrac{1}{18}\)

Kết hợp ĐKXĐ, ta được: \(0\le x< \dfrac{1}{18}\)

b: Ta có: \(\sqrt{-3x+\dfrac{1}{2}}\ge5\)

\(\Leftrightarrow-3x+\dfrac{1}{2}\ge25\)

\(\Leftrightarrow-3x\ge\dfrac{49}{2}\)

hay \(x\le-\dfrac{49}{6}\)

c: Ta có: \(\sqrt{-2x+1}>7\)

\(\Leftrightarrow-2x+1>49\)

\(\Leftrightarrow-2x>48\)

hay x<-24