\(\left(x-2\right)^2+\dfrac{24}{x^2}+\dfrac{22}{x}=11\left(x\ne0\right)\)
\(\Leftrightarrow x^2\left(x-2\right)^2+24+22x=11x^2\)
\(\Leftrightarrow x^2\left(x^2-4x+4\right)+22x+24=11x^2\)
\(\Leftrightarrow x^4-4x^3+4x^2+22x+24-11x^2=0\)
\(\Leftrightarrow x^4-4x^3-7x^2+22x+24=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\\x-3=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=3\\x=4\end{matrix}\right.\) (Thỏa mãn)
Vậy nghiệm của phương trình cho là \(x\in\left\{-2;-1;3;4\right\}\)
ĐKXĐ: x<>0
Ta có: \(\left(x-2\right)^2+\dfrac{24}{x^2}+\dfrac{22}{x}=11\)
=>\(\left(x-2\right)^2-4+\dfrac{24}{x^2}-1,5+\dfrac{22}{x}-5,5=0\)
=>\(\left(x-2-2\right)\left(x-2+2\right)+1,5\left(\dfrac{16}{x^2}-1\right)+\dfrac{11}{2}\left(\dfrac{4}{x}-1\right)=0\)
=>\(\left(x-4\right)\cdot x+1,5\left(\dfrac{4}{x}-1\right)\left(\dfrac{4}{x}+1\right)+\dfrac{11}{2}\left(\dfrac{4}{x}-1\right)=0\)
=>\(\left(x-4\right)\left[x+1,5\cdot\dfrac{4+x}{x}\cdot\dfrac{1}{x}+\dfrac{11}{2}\cdot\dfrac{1}{x}\right]=0\)
=>\(\left(x-4\right)\left(x+\dfrac{6+1,5x}{x^2}-\dfrac{11}{2x}\right)=0\)
=>\(\left(x-4\right)\left(\dfrac{2x^3+12+3x-11x}{2x^2}\right)=0\)
=>\(\left(x-4\right)\left(2x^3-8x+12\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\2x^3-8x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x\simeq-2,53\left(nhận\right)\end{matrix}\right.\)