Question

giải các phương trình sau:

a) \(\sqrt{4x^2-9}=\sqrt{2x-3}\)

b) \(\sqrt{x^2-1}-\sqrt{x-1}=0\)

Answer 1

a: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=1\end{matrix}\right.\Leftrightarrow x=\dfrac{3}{2}\)

Answer 2

\(a,ĐK:x\ge\dfrac{3}{2}\\ PT\Leftrightarrow\sqrt{2x-3}\left(\sqrt{2x+3}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\\sqrt{2x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{3}{2}\)

\(b,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=1\)