Question

Tính : \(\frac{\sqrt{x+1}}{\sqrt{x-5}}=2\)

Answer 1

ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ge0\\x-5>0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge-1\\x>5\end{matrix}\right.\)

=> \(x>5\)

Ta có : \(\frac{\sqrt{x+1}}{\sqrt{x-5}}=2\)

<=> \(\sqrt{x+1}=2\sqrt{x-5}\)

<=> \(\left(\sqrt{x+1}\right)^2=\left(2\sqrt{x-5}\right)^2\)

<=> \(x+1=4\left(x-5\right)\)

<=> \(x+1=4x-20\)

<=> \(20+1=4x-x\)

<=> \(3x=21\)

<=> \(x=7\left(TM\right)\)

Vậy phương trình trên có nghiệm là x = 7 .