1 câu trả lời
#LC
`\sqrt(x^2-1) - 2\sqrt(x+1)=0` ĐKXĐ: \(\left[ \begin{array}{l}x\geq1\\x\leq-1\end{array} \right.\)
⇔ `\sqrt((x-1)(x+1)) - 2\sqrt(x+1)=0`
⇔ `\sqrt(x-1).\sqrt(x+1) - 2\sqrt(x+1)=0`
⇔ `\sqrt(x+1).(\sqrt(x-1)-2) =0`
⇔ \(\left[ \begin{array}{l}\sqrt{x+1}=0\\\sqrt{x-1}=2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x+1=0\\x-1=4\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-1(tm)\\x=5(tm)\end{array} \right.\)
Vậy, `x ∈ {-1;5}`