\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
\(pt\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}-2\left(x+1\right)=0\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+1}=0\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}-2\sqrt{x+1}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\end{matrix}\right.\)
Xét \(\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow2\left(x+3\right)+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\)
\(\Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\)
\(\Leftrightarrow8\left(x+3\right)\left(x-1\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x+25\right)=0\Rightarrow x=1\) (Thỏa)
Vậy nghiệm của pt là \(x=±1\)
\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
\(\Leftrightarrow\sqrt{2\left(x+1\right)\left(x+3\right)}+\sqrt{\left(x-1\right)\left(x+1\right)}=2\left(x+1\right)\)
\(\Leftrightarrow\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow2\left(x+3\right)+\left(x-1\right)+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\)
\(\Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\)
\(\Leftrightarrow\sqrt{x-1}\left(2\sqrt{2x+6}-\sqrt{x-1}\right)=0\)
TH1:
x - 1 = 0
<=> x = 1 (nhận)
TH2:
\(2\sqrt{2x+6}=\sqrt{x-1}\)
\(\Leftrightarrow4\left(2x+6\right)=x-1\)
\(\Leftrightarrow x=-\dfrac{25}{7}\) (loại)
Vậy x = 1
