Đặt \(t=3x^2+5x+2\)
Do đó ta có:\(\sqrt{3x^2+5x+7}-\sqrt{3x^2+5^2+2}=1\)
\(\sqrt{t+5}-\sqrt{t}=1\)
\(\left(\sqrt{t+5}-\sqrt{t}\right)^2=1\)
\(t+5-2\sqrt{t\left(t+5\right)}+t=1\)
\(2t-2\sqrt{t\left(t+5\right)}+5=1\)
\(2t+4=2\sqrt{t\left(t+5\right)}\)
\(\left(t+2\right)^2=t\left(t+5\right)\)
\(4t+4=5t\)
\(\Rightarrow t=4\)
Tại t=4 ta được:\(3x^2+5x+2=4\)
\(3x^2+5x-2=0\)
\(3x^2+6x-x-2=0\)
\(\Rightarrow\left(3x-1\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)
