Ta có: \(\sqrt{5x^2+6x+5}=\dfrac{64x^3+4x}{5x^2+6x+6}\)
\(\Leftrightarrow\sqrt{5x^2+6x+5}-4x=\dfrac{64x^3+4x}{5x^2+6x+6}-4x\)
\(\Leftrightarrow-\dfrac{\left(x-1\right)\left(11x+5\right)}{\sqrt{5x^2+6x+5+4x}}=\dfrac{4x\left(11x+5\right)\left(x-1\right)}{5x^2+6x+6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,x=-\dfrac{5}{11}\left(loại\right)\\\dfrac{-1}{\sqrt{5x^2+6x+5+4x}}=\dfrac{4x}{5x^2+6x+6}\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất \(x=1\)
\(\Leftrightarrow\left(5x^2+6x+6\right)\sqrt{5x^2+6x+5}=64x^3+4x\)
\(\Leftrightarrow\left(5x^2+6x+5\right)\sqrt{5x^2+6x+5}+\sqrt{5x^2+6x+5}=64x^3+4x\)
Đặt \(\left\{{}\begin{matrix}\sqrt{5x^2+6x+5}=a>0\\4x=b\end{matrix}\right.\)
\(\Rightarrow a^3+a=b^3+b\)
\(\Rightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+a-b=0\)
\(\Rightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Rightarrow\left(a-b\right)\left[\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}+1\right]=0\)
\(\Rightarrow a=b\)
\(\Rightarrow\sqrt{5x^2+6x+5}=4x\left(x\ge0\right)\)
\(\Rightarrow5x^2+6x+5=16x^2\)
\(\Rightarrow11x^2-6x-5=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{11}< 0\left(loại\right)\end{matrix}\right.\)
