ĐKXĐ: x<>1
Đặt a=x; \(b=\frac{x}{x-1}\)
\(a+b=x+\frac{x}{x-1}=\frac{x^2-x+x}{x-1}=\frac{x^2}{x-1}\)
\(ab=x\cdot\frac{x}{x-1}=\frac{x^2}{x-1}\)
TA có: \(x^3+\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2}{x-1}=2\)
=>\(a^3+b^3+3ab=2\)
=>\(\left(a+b\right)^3-3ab\left(a+b\right)+3ab=2\)
=>\(\left(\frac{x^2}{x-1}\right)^3-3\cdot\frac{x^2}{x-1}\cdot\frac{x^2}{x-1}+3\cdot\frac{x^2}{x-1}=2\)
=>\(\left(\frac{x^2}{x-1}\right)^3-3\cdot\left(\frac{x^2}{x-1}\right)^2+3\cdot\frac{x^2}{x-1}-1=1\)
=>\(\left(\frac{x^2}{x-1}-1\right)^3=1\)
=>\(\frac{x^2}{x-1}-1=1\)
=>\(\frac{x^2}{x-1}=2\)
=>\(x^2=2x-2\)
=>\(x^2-2x+2=0\)
=>\(\left(x-1\right)^2+1=0\) (vô lý)
=>x∈∅
