ĐKXĐ : \(x\ne\pm3\)
Với \(x\ne\pm3\) ta có :
pt \(\Leftrightarrow\frac{2\left(x+3\right)}{x^2-9}+\frac{3\left(x-3\right)}{x^2-9}=\frac{3x+5}{x^2-9}\)
\(\Leftrightarrow2x+6+3x-9=3x+5\)
\(\Leftrightarrow5x-3=3x+5\)
\(\Leftrightarrow2x=8\Leftrightarrow x=4\) ( TM ĐKXĐ )
vậy x = 4 là nghiệm duy nhất của pt đã cho
\(\frac{2}{x-3}\)+\(\frac{3}{x+3}\)=\(\frac{3x+5}{x^{2^{ }}-9}\) :DKXD x\(\ne\)3;x\(\ne\)-3
\(\Leftrightarrow\)\(\frac{2}{x-3}\)+\(\frac{3}{x+3}\)=\(\frac{3x+5}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\)\(\frac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\)+\(\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)=\(\frac{3x+5}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\)2x+6+3x-9=3x+5
\(\Leftrightarrow\)2x+3x-3x=5+9-6
\(\Leftrightarrow\)2x=8
\(\Leftrightarrow\)x=4(TMDKXD)
vay s={4}
chuc bn hoc tot
