ĐKXĐ:\(x\ne\pm2\)
\(\dfrac{6}{x-2}+\dfrac{5}{x+2}=\dfrac{x-18}{x^2-4}\\ \Leftrightarrow\dfrac{6\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{x-18}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{6x+12+5x-10-x+18}{\left(x+2\right)\left(x-2\right)}=0\\ \Rightarrow10x+20=0\\ \Leftrightarrow x=-2\left(ktm\right)\)
\(\dfrac{6}{x-2}-\dfrac{5}{x+2}=\dfrac{x-18}{x}\)
mũ 4 ở đâu vậy ạ
\(\dfrac{6}{x-2}+\dfrac{5}{x+2}=\dfrac{18}{x^2-2^2}\)
\(\dfrac{6.\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{5\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\dfrac{18}{\left(x-2\right)\left(x+2\right)}\)
khử mẫu
=>6x+12+5x-10=18-x
=>6x+12+5x-10-18+x=0
=>10x-20=0
=>x=2(loại)
=> pt vô nghiện
