Question

\(\dfrac{3}{x^2-3x}\)+\(\dfrac{1}{x}\)=\(\dfrac{x+4}{x-3}\)

Answer 1

ĐKXĐ: \(x\ne0;x\ne3\)

\(\dfrac{3}{x^2-3x}+\dfrac{1}{x}=\dfrac{x+4}{x-3}\)

⇔ \(\dfrac{3}{x\left(x-3\right)}+\dfrac{x-3}{x\left(x-3\right)}=\dfrac{x\left(x+4\right)}{x\left(x-3\right)}\)

⇒ \(3+x-3=x^2+4x\)

⇔ \(x^2-3x=0\)

⇔ \(x\left(x-3\right)=0\)

⇒ \(\left\{{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)  ⇒ \(\left\{{}\begin{matrix}x=0\left(K^OTMĐK\right)\\x=3\left(K^OTMĐK\right)\end{matrix}\right.\)

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