\(2x^2+2\left(x+1\right)\sqrt{x^2-1}=6x-1\\ \leftrightarrow2\left(x+1\right)\sqrt{x^2-1}=6x-1-2x^2\\ \leftrightarrow4\left(x+1\right)^2\left(x^2-1\right)=36x^2+1+4x^4-12x-24x^3+4x^2\\ \leftrightarrow8x^3-4x^2-8x-4=36x^2+1-12x-24x^3\\ \leftrightarrow8x^3-4x^2-8x-4-36x^2-1+12x+24x^3=0\\ \leftrightarrow32x^3-40x^2+4x-5=0\\ \leftrightarrow8x^2\left(4x-5\right)+4x-5=0\\ \leftrightarrow\left(4x-5\right)\left(8x^2+1\right)=0\\ \leftrightarrow\left[{}\begin{matrix}4x-5=0\\8x^2+1=0\end{matrix}\right.\leftrightarrow\left[{}\begin{matrix}4x=5\\8x^2+1=0\end{matrix}\right.\leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\left(TM\right)\\8x^2+1=0\left(VN\right)\end{matrix}\right.\)
Vậy...
đặt x+1=a ;x-1=b
\(\Leftrightarrow2ab+2+2a\sqrt[]{ab}=6b-5\)
