\(\sqrt{2x-1}\)=3
=2x-1=3
=2x=3+1
=2x=4
x=\(\dfrac{4}{2}\)
x=2
vậy x =2
sữa dê 1 chút nhé \(\sqrt{7x}\)-\(\sqrt{2x-3}\)=0
= 7x-2x-3=0
= 5x -3=0
= 5x =3
x= \(\dfrac{3}{5}\)
vậy x = \(\dfrac{3}{5}\)
\(\dfrac{3}{5}\)
\(\sqrt{2x-1}=3\Leftrightarrow2x-1=9\Leftrightarrow2x=10\Leftrightarrow x=5\)
Có thể thử lại bằng MT
\(\sqrt{7x}-\sqrt{2x-3}=0\)
\(\Leftrightarrow\sqrt{7x}=\sqrt{2x-3}\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\7x=2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\5x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x=\dfrac{-3}{5}\end{matrix}\right.\)
(Vô lí )
Vậy pt vô nghiệm.
