ĐKXĐ: \(x\ge-\dfrac{2}{5}\)
\(\sqrt{5x+2}=\sqrt{x+3}\)
\(\Rightarrow5x+2=x+3\)
\(\Rightarrow4x=1\)
\(\Rightarrow x=\dfrac{1}{4}\) (thỏa mãn)
\(\sqrt{5x+2}-\sqrt{x+3}=0\)
<=> \(\sqrt{5x+2}=\sqrt{x+3}\)
<=> 5x + 2 = x + 3
<=> 5x - x = 3 - 2
<=> 4x = 1
<=> x = \(\dfrac{1}{4}\)
Ta có: \(\sqrt{5x+2}-\sqrt{x+3}=0\)
\(\Leftrightarrow5x+2=x+3\)
hay \(x=\dfrac{1}{4}\)