ĐKXĐ: \(\orbr{\begin{cases}x\le-4-\sqrt{7}\\x\ge-1\end{cases}}\)
Để phương trình có nghiệm thì \(2x+4\ge0\Leftrightarrow x\ge-2\)
Ta có:
\(\sqrt{2x^2+16x+18}+\sqrt{x^2-1}=2x+4\)
\(\Leftrightarrow\sqrt{\left(4x^2+16x+16\right)-2\left(x^2-1\right)}+\sqrt{x^2-1}=2x+4\)
\(\Leftrightarrow\sqrt{\left(2x+4\right)^2-2\left(x^2-1\right)}+\sqrt{x^2-1}=2x+4\)
Đặt \(\hept{\begin{cases}\sqrt{x^2-1}=a\\2x+4=b\end{cases}}\left(a,b\ge0\right)\)
Ta có: \(\sqrt{b^2-2a^2}+a=b\)
\(\Leftrightarrow\sqrt{b^2-2a^2}=b-a\)
\(\Leftrightarrow\hept{\begin{cases}b\ge a\\b^2-2a^2=b^2-2ab+a^2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b\ge a\\3a^2-2ab=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b\ge a\\a\left(3a-2b\right)=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}a=0\\a=\frac{2}{3}b\end{cases}}\)
TH1: \(a=0\Leftrightarrow\sqrt{x^2-1}=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
Thử lại x = 1 hoặc x = - 1 thỏa mãn.
TH2: \(a=\frac{2}{3}b\Leftrightarrow\sqrt{x^2-1}=\frac{2}{3}\left(2x+4\right)\)
\(\Leftrightarrow9x^2-9=16x^2+64x+64\)
\(\Leftrightarrow7x^2+64x+73=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-32+3\sqrt{57}}{7}\left(tm\right)\\x=\frac{-32-3\sqrt{57}}{7}\left(l\right)\end{cases}}\)
Vậy pt có 3 nghiệm \(x=1;x=-1;x=\frac{-32+3\sqrt{57}}{7}\)
