a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge-1\\x>1\end{matrix}\right.\)\(\Rightarrow x>1\)
Ta có : \(PT\Leftrightarrow\sqrt{x+1}=2\sqrt{x-1}\)
\(\Leftrightarrow x+1=4x-4\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\left(TM\right)\)
Vậy ...
b, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge1\\x>-1\end{matrix}\right.\)\(\Rightarrow x\ge1\)
Ta có : \(PT\Leftrightarrow\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow x-1=4x+4\)
\(\Leftrightarrow3x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{3}\left(L\right)\)
Vậy phương trình vô nghiệm .
a) ĐKXĐ: \(x>1\)
Ta có: \(\dfrac{\sqrt{x+1}}{\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x+1}=2\sqrt{x-1}\)
\(\Leftrightarrow x+1=4x-4\)
\(\Leftrightarrow x-4x=-4-1\)
\(\Leftrightarrow-3x=-5\)
hay \(x=\dfrac{5}{3}\left(nhận\right)\)
Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x>-1\\x\ne1\end{matrix}\right.\)
Ta có: \(\dfrac{\sqrt{x-1}}{\sqrt{x+1}}=2\)
\(\Leftrightarrow\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow x-1=4x+4\)
\(\Leftrightarrow x-4x=4+1\)
\(\Leftrightarrow-3x=5\)
hay \(x=-\dfrac{5}{3}\)(loại)
Vậy: \(S=\varnothing\)
