\(\frac{x+2}{x-3}-\frac{x-5}{x+3}=\)\(\frac{2x}{x^2-9}\) (ĐKXĐ : x\(\ne\pm3\))
\(\Leftrightarrow\frac{x+2}{x-3}-\frac{x-5}{x+3}=\frac{2x}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\)\(\frac{\left(x+2\right)\left(x+3\right)-\left(x-5\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2x}{\left(x+3\right)\left(x-3\right)}\)
Suy ra :
\(\Leftrightarrow\)(x+2)(x+3)-(x-5)(x-3)=2x
\(\Leftrightarrow\)x2+3x+2x+6-x2+3x+5x-15 =2x
\(\Leftrightarrow\)x2+3x+2x-x2+3x+5x = 15-6
\(\Leftrightarrow\)3x =9
\(\Leftrightarrow\)x = 3 (tm)
Vậy phương trình có tập nghiệm S = \(\left\{3\right\}\)
\(\Leftrightarrow\)
mik nhầm phần cuối nhé x=3 (ko tm)
\(\Rightarrow\)pt vô nghiệm
