ĐKXĐ: x<>0
Ta có: \(\left(x-2\right)^2+\dfrac{24}{x^2}+\dfrac{22}{x}=11\)
=>\(\left(x-2\right)^2-4+\dfrac{24}{x^2}-1,5+\dfrac{22}{x}-5,5=0\)
=>\(\left(x-2-2\right)\left(x-2+2\right)+1,5\left(\dfrac{16}{x^2}-1\right)+\dfrac{11}{2}\left(\dfrac{4}{x}-1\right)=0\)
=>\(\left(x-4\right)\cdot x+1,5\left(\dfrac{4}{x}-1\right)\left(\dfrac{4}{x}+1\right)+\dfrac{11}{2}\left(\dfrac{4}{x}-1\right)=0\)
=>\(\left(x-4\right)\left[x+1,5\cdot\dfrac{4+x}{x}\cdot\dfrac{1}{x}+\dfrac{11}{2}\cdot\dfrac{1}{x}\right]=0\)
=>\(\left(x-4\right)\left(x+\dfrac{6+1,5x}{x^2}-\dfrac{11}{2x}\right)=0\)
=>\(\left(x-4\right)\left(\dfrac{2x^3+12+3x-11x}{2x^2}\right)=0\)
=>\(\left(x-4\right)\left(2x^3-8x+12\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\2x^3-8x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x\simeq-2,53\left(nhận\right)\end{matrix}\right.\)
