Question

BT : Tìm Điều kiện của phương trình

a, \(\sqrt{2x+1}=\dfrac{1}{x}\)
b,
\(\dfrac{x+2}{\sqrt{2x^2}+1}=\dfrac{2}{\sqrt{x+3}}\)

c,\(\dfrac{x}{\sqrt{x-1}}=\dfrac{2}{\sqrt{x}+3}\)

d , \(\dfrac{2x+3}{x^2-4}=\sqrt{x+1}\)

Answer 1

a) đk \(\left\{{}\begin{matrix}2x+1\ge0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\ne0\end{matrix}\right.\)

b) đk \(x+3>0\Leftrightarrow x>-3\)

c) \(\left\{{}\begin{matrix}x-1>0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x\ge0\end{matrix}\right.\Leftrightarrow x>1\)

d) đk \(\left\{{}\begin{matrix}x^2-4\ne0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne\pm2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne2\end{matrix}\right.\)