Question

1. \(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)

2. \(\sqrt{4x^2-9}=3\sqrt{2x-3}\)

Giải phương trình

huhu giúp e với e đang cần gấp ạa :(( e cảm ơnnn

Answer 1

1)

ĐKXĐ: x>4

Ta có: \(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)

\(\Leftrightarrow x^2+8x+15=x^2-6x+8\)

\(\Leftrightarrow8x+6x=8-15\)

\(\Leftrightarrow14x=-7\)

hay \(x=-\dfrac{1}{2}\)(loại)

2) Ta có: \(\sqrt{4x^2-9}=3\sqrt{2x-3}\)

\(\Leftrightarrow\sqrt{2x-3}\left(\sqrt{2x+3}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)