Ta có: \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)

<=> \(\left(x^2+x\right)\left(x+6\right)-x^3=5x\)

<=> \(x^3+7x^2+6x-x^3=5x\)

<=> \(7x^2+x=0\)

<=> \(x\left(7x+1\right)=0\)

<=> \(\left[\begin{array}{nghiempt}x=0\\7x+1=0\end{array}\right.\)<=>\(\left[\begin{array}{nghiempt}x=0\\x=-\frac{1}{7}\end{array}\right.\)

Vậy x\(\in\left\{-\frac{1}{7};0\right\}\)

Ta có: \(f\left(x\right)=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)= \(\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}{\left(\sqrt{x+1}-\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}\)=\(\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{x+1-\left(x-1\right)}\)

                     = \(\frac{x+1+x-1+2\sqrt{\left(x-1\right)\left(x+1\right)}}{2}\)= \(\frac{2x+2\sqrt{x^2-1}}{2}\)=\(x+\sqrt{x^2-1}\)

Với a= \(\sqrt{3}\)=> \(f\left(\sqrt{3}\right)=\sqrt{3}+\sqrt{\left(\sqrt{3}\right)^2-1}\)=\(\sqrt{3}+\sqrt{2}\)