ĐKXĐ : x khác 49 , x khác 50
Ta có :
\(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)
\(\Leftrightarrow\frac{x-49}{50}-1+\frac{x-50}{49}-1=\frac{49}{x-50}-1+\frac{50}{x-49}-1\)
\(\Leftrightarrow\frac{x-99}{50}+\frac{x-99}{49}=\frac{99-x}{x-50}+\frac{99-x}{x-49}\)
\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{x-50}+\frac{1}{x-49}\right)=0\)
Mà 1/50 + 1/49 + 1/x-50 + 1/x-49 khác 0
\(\Leftrightarrow x-99=0\)
\(\Leftrightarrow x=99\)
@Kyo-kun
\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{50}+\frac{1}{x-50}+\frac{1}{49}+\frac{1}{x-49}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-99=0\\\frac{x}{50\left(x-50\right)}+\frac{x}{49\left(x-49\right)}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=99\\x\left(\frac{1}{50\left(x-50\right)}+\frac{1}{49\left(x-49\right)}\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=99\\x=0\end{cases}\left(t.m\right)}}\)
Vậy x = 99 hoặc x = 0
