\(\left(x+2\right)\left(\sqrt{2x+3}-2\sqrt{x+1}\right)+\sqrt{2x^2+5x+3}-1=0\left(-1\le x\right)\\ \Leftrightarrow\left(x+2\right)\left(\sqrt{2x+3}-2\sqrt{x+1}\right)+\sqrt{\left(2x+3\right)\left(x+1\right)}-1=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\\\sqrt{x+1}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2-b^2=x+2\\a-2b=1\end{matrix}\right.\), pt trở thành:
\(\left(a^2-b^2\right)\left(a-2b\right)+ab-a^2+2b^2=0\\ \Leftrightarrow\left(a-2b\right)\left(a-b\right)\left(a+b\right)-\left(a-b\right)\left(a-2b\right)=0\\ \Leftrightarrow\left(a-2b\right)\left(a-b\right)\left(a+b-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\a=b\\a+b-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=2\sqrt{x+1}\\\sqrt{2x+3}=\sqrt{x+1}\\\sqrt{2x+3}+\sqrt{x+1}-1=0\end{matrix}\right.\)
\(x\ge1\Leftrightarrow a\ge1;b\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4x+4\\2x+3=x+1\\\sqrt{2x+3}+\sqrt{x+1}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{7}\left(ktm\right)\\x=-2\left(ktm\right)\\\sqrt{2x+3}+\sqrt{x+1}-1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)=a+b-1\ge1+0-1=0\)
Dấu \("="\Leftrightarrow x=-1\)
Vậy pt có nghiệm duy nhất là \(x=-1\)
