Question

Answer 1

\(a,3\sqrt{x}-1=0\left(x\ge0\right)\\ \Leftrightarrow\sqrt{x}=\dfrac{1}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)

\(b,\sqrt{6x-1}=\sqrt{35}\left(x\ge\dfrac{1}{6}\right)\\ \Leftrightarrow6x-1=35\\ \Leftrightarrow x=6\)

 

Answer 2

a. ĐK: \(x\ge0\)
\(3\sqrt{x}-1=0\)

\(\Rightarrow3\sqrt{x}=1\)

\(\Rightarrow\sqrt{x}=\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{1}{9}\)

b. \(\sqrt{6x-1}=\sqrt{35}\)

\(\Rightarrow6x-1=35\)

\(\Rightarrow6x=36\)

\(\Rightarrow x=6\)

Answer 3

a: Ta có: \(3\sqrt{x}-1=0\)

\(\Leftrightarrow3\sqrt{x}=1\)

hay \(x=\dfrac{1}{9}\)

b: Ta có: \(\sqrt{6x-1}=\sqrt{35}\)

\(\Leftrightarrow6x-1=35\)

\(\Leftrightarrow6x=36\)

hay x=6