\(2\sqrt{x+1}-\sqrt{x+6}=1\)
ĐKXĐ : x\(\ge-1\)
\(\Leftrightarrow2\sqrt{x+1}=1+\sqrt{x+6}\)
bình phương 2 vế ta có :
\(4x+4=1+2\sqrt{x+6}+x+6\)
\(3x-3=2\sqrt{x+6}\)
ĐK : \(x>1\)
bình phương 2 vế ta có :
\(9x^2-18x+9=4x+24\Leftrightarrow9x^2-22x-23=0\)\(\Leftrightarrow9x^2-2.11.x+121-144=0\)
\(\Leftrightarrow\left(3x-11\right)^2=144\Leftrightarrow3x-11=\pm12\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-11=12\\3x-11=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
kết hợp vs đkxđ => x=23/3