\(\frac{3}{x-3}+\frac{1}{x+3}=\frac{6}{x^2-9}\) đkxđ \(x\ne\pm3\)
\(\Leftrightarrow\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow3x+9+x+3=6\)
\(\Leftrightarrow3x+x=6-3-9\)
\(\Leftrightarrow4x=-6\)
\(\Leftrightarrow x=-\frac{3}{2}\)
\(\dfrac{3}{{x - 3}} + \dfrac{1}{{x + 3}} = \dfrac{6}{{{x^2} - 9}}\)
ĐK: \(x \ne3;x\ne-3\)
\(\left( {x + 3} \right) + x - 3 = 6\\ \Leftrightarrow 3x + 9 + x - 3 = 6\\ \Leftrightarrow 4x = 0\\ \Leftrightarrow x = 0\left( {tm} \right) \)
Vậy $x=0$ là nghiệm phương trình
\(\frac{3}{x-3}+\frac{1}{x+3}=\frac{6}{x^2-9}\) \(\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow3\left(x+3\right)+x-3=6\)
\(\Leftrightarrow4x+6=6\Leftrightarrow x=0\)
Vậy phương trình có nghiệm x = 0.
\(\frac{3}{x-3}+\frac{1}{x+3}=\frac{6}{x^2-9}ĐKXĐ:x\ne\pm3\)
\(\frac{3}{x-3}+\frac{1}{x+3}=\frac{6}{\left(x+3\right)\left(x-3\right)}\)
\(3\left(x+3\right)+x-3=6\)
\(3x+9+x-3=6\)
\(4x+6=6\)
\(4x=0\)
\(x=0\)
