Question

[phanhuy]

giải phương trình x/(x+3)+6/(x-3)=-18/(9-x^2)

Answer 1

[乇尺尺のﾚ]

\(\dfrac{x}{x+3}+\dfrac{6}{x-3}=\dfrac{-18}{9-x^2}\)

\(\Leftrightarrow\dfrac{x}{x+3}+\dfrac{6}{x-3}=\dfrac{18}{x^2-9}\)

\(ĐKXĐ:\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm3\)

\(\dfrac{x}{x+3}+\dfrac{6}{x-3}=\dfrac{18}{x^2-9}\)

\(\Leftrightarrow\dfrac{x.\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{6.\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{18}{\left(x+3\right)\left(x-3\right)}\)

\(\Rightarrow x^2-3x+6x+18=18\)

\(\Leftrightarrow x^2-3x+6x=18-18\)

\(\Leftrightarrow x^2+3x=0\)

\(\Leftrightarrow x\left(x+3\right)=0\)

\(\Leftrightarrow x=0hoặcx+3=0\)

\(\Leftrightarrow x=0\left(tm\right)hoặcx=-3\left(ktm\right)\)

Vậy phương trình có nghiệm là \(x=0\)