ĐKXĐ: x<>1
Ta có: \(\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2}{x-1}=2\)
=>\(\frac{x^3}{\left(x-1\right)^3}+\frac{3x^2\left(x-1\right)^2}{\left(x-1\right)^3}=2\)
=>\(x^3+3x^2\left(x-1\right)^2=2\left(x-1\right)^3\)
=>\(x^2\left\lbrack x+3\left(x-1\right)^2\right\rbrack=2\left(x-1\right)^3\)
=>\(x^2\left(3x^2-6x+3+x\right)=2\left(x^3-3x^2+3x-1\right)\)
=>\(3x^4-5x^3+3x^2-2x^3+6x^2-6x+2=0\)
=>\(3x^4-7x^3+9x^2-6x+2=0\)
=>\(\left(x^2-x+1\right)\left(3x^2-4x+2\right)=0\)
mà \(x^2-x+1=\left(x-\frac12\right)^2+\frac34>0\forall x\)
nên \(3x^2-4x+2=0\)
=>\(x^2-\frac43x+\frac23=0\)
=>\(x^2-2\cdot x\cdot\frac23+\frac49+\frac29=0\)
=>\(\left(x-\frac23\right)^2+\frac29=0\) (vô lý)
=>x∈∅
