\(\sqrt{9x+9}-3\sqrt{x+1}-\sqrt{4x+4}=1\)
ĐK : x ≥ -1
\(\Leftrightarrow\sqrt{3^2\left(x+1\right)}-3\sqrt{x+1}-\sqrt{2^2\left(x+1\right)}=1\)
\(\Leftrightarrow3\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=1\)
\(\Leftrightarrow-2\sqrt{x+1}=1\)
<=> \(\sqrt{x+1}=-\frac{1}{2}\)( dễ thấy điều này vô lí vì \(\sqrt{x+1}\ge0\forall x\ge-1\))
=> pt vô nghiệm
\(3\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=1\)
\(\left(3-3-2\right)\sqrt{x+1}=1\)
\(-2\sqrt{x+1}=1\)
\(\sqrt{x+1}=\frac{-1}{2}\)
\(x+1=\left(\frac{-1}{2}\right)^2\)
\(x=\frac{1}{4}-1=\frac{-3}{4}\)
Vậy x =-3/4
