a) ĐK:\(x\ge\dfrac{1}{2}\)
\(\sqrt{2x-1}=\sqrt{5}\Leftrightarrow\left(\sqrt{2x-1}\right)^2=\left(\sqrt{5}\right)^2\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\left(tm\right)\)
Vậy S={3}
b) ĐK:\(x\ge5\)
\(\sqrt{x-5}=3\Leftrightarrow\left(\sqrt{x-5}\right)^2=3^2\Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)
Vậy S={14}
a,\(ĐK:\sqrt{2x-1}\ge0\Leftrightarrow x\ge\dfrac{1}{2}
\)
Ta có \(\sqrt{2x-1}=\sqrt{5}\)
\(\Rightarrow2x-1=5\)
\(\Leftrightarrow2x=6\Leftrightarrow x=3\)(tm)
Vậy nghiệm của pt S=\(\left\{3\right\}\)
b,ĐK \(\sqrt{x-5}\ge0\Leftrightarrow x\ge5\)
Ta có: \(\sqrt{x-5}=3\)
\(\Rightarrow x-5=9\)
\(\Leftrightarrow x=14\)(tm)
Vậy nghiệm của pt S=\(\left\{14\right\}\)