mk nghĩ bạn chép sai đề hình như đề bài phải là \(A=\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\)

ta xét \(A^3=\left(\sqrt[3]{\frac{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}}{2}}+\sqrt[3]{\frac{x^3-3x-\left(x^2-1\right)\sqrt{x^2-4}}{2}}\right)^3\)

  <=> \(A^3=x^3-3x+3A\cdot\sqrt[3]{\frac{4}{4}}\)

<=> \(A^3=x^3-3x+3A\)

<=> \(A^3-3A-x^3+3x=0\)

<=>\(\left(A^3-x^3\right)-3A+3x=0\)

<=> \(\left(A-x\right)\left(A^2+Ax+x^2\right)-3\left(A-x\right)=0\)

<=> \(\left(A-x\right)\left(A^2+Ax+x^2-3\right)=0\)

<=> \(\orbr{\begin{cases}A=x\\A^2+Ax+x^2-3=0\end{cases}}\)(vô lí )

vậy \(A=x\)

Ta có: 

A = \(\frac{1x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}-2}{x^3-3x+\left(x^2-1\right)\sqrt{x^2-4}-2}\)

= \(\frac{\left(x-2\right)\left(x+1\right)^2+\left(x^2-1\right)\sqrt{x^2-4}}{\left(x+2\right)\left(x-1\right)^2+\left(x^2-1\right)\sqrt{x^2-4}}\)

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

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

Ủa cái này thì A=1 mà

xin lỗi, viết nhầm, trên tử là -2

x + 254 = 245 + 368

x + 254 = 613

x           = 613 - 254

x           = 359