Câu hỏi của Jum Võ

Question

$\dfrac{1}{x^2-2x+2}+\dfrac{2}{x}=\dfrac{3}{x^2+2x+2}.$

Nguyễn Việt Lâm · Accepted Answer

ĐKXĐ: $x\ne0$

$\dfrac{1}{x^2-2x+2}+\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{3}{x^2+2x+2}=0$

$\Leftrightarrow\dfrac{x^2-x+2}{x\left(x^2-2x+2\right)}+\dfrac{x^2-x+2}{x\left(x^2+2x+2\right)}=0$

$\Leftrightarrow\dfrac{x^2-x+2}{x}\left(\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2+2x+2}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2-x+2}{x}=0\left(vn\right)\\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2+2x+2}=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x^2-x+2=0\left(vn\right)\2x^2+4=0\left(vn\right)\end{matrix}\right.$

$\Rightarrow$ Phương trình vô nghiệmCâu trả lời: ĐKXĐ: $x\ne0$

$\dfrac{1}{x^2-2x+2}+\dfrac{1}{x}+\dfrac{1}{x}-\dfrac{3}{x^2+2x+2}=0$

$\Leftrightarrow\dfrac{x^2-x+2}{x\left(x^2-2x+2\right)}+\dfrac{x^2-x+2}{x\left(x^2+2x+2\right)}=0$

$\Leftrightarrow\dfrac{x^2-x+2}{x}\left(\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2+2x+2}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2-x+2}{x}=0\left(vn\right)\\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2+2x+2}=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x^2-x+2=0\left(vn\right)\2x^2+4=0\left(vn\right)\end{matrix}\right.$

$\Rightarrow$ Phương trình vô nghiệm

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