dkxd:\(x\ne0;x\ne1;x\ne2;x\ne3;x\ne4\)
mình sửa để lại 1 tí nhé, để bạn sai rồi
\(\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}=2+\dfrac{1}{4-x}\\ \Leftrightarrow\dfrac{1}{x^2-1}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}=2-\dfrac{1}{x-4}\\ \Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}+\dfrac{1}{x-2}+\dfrac{1}{x-2}+\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}=2-\dfrac{1}{x-4}\\ \Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x-4}-2+\dfrac{1}{x-4}=0\\ \Leftrightarrow\dfrac{x-4-x-2x\left(x-4\right)+x}{x\left(x-4\right)}=0\\ \Leftrightarrow x-4-x-2x^2+8x+x=0\\ \Leftrightarrow-2x^2+9x-4=0\\ \Leftrightarrow-\left(2x^2-9x+4\right)=0\\ \Leftrightarrow-\left(2x^2-8x-x+4\right)=0\\ \Leftrightarrow-\left(2x\left(x-4\right)-\left(x-4\right)\right)=0\\ \Leftrightarrow-\left(x-4\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(ktmdkxd\right)\\x=\dfrac{1}{2}\left(tmdkxd\right)\end{matrix}\right.\)
Vậy x=1/2
bạn trần thị ngọc trâm làm sai chỗ 1/x^2-x rồi bạn pân tích phải la x(x-1)
