Đề sao có tận ba dấu bằng thế bạn?
*Sửa đề: \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)
ĐKXĐ: x≠1;x≠-3
Ta có: \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow2x\left(x+3\right)+4-\left(2x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow2x^2+6x+4-\left(2x^2-2x-5x+5\right)=0\)
\(\Leftrightarrow2x^2+6x+4-\left(2x^2-7x+5\right)=0\)
\(\Leftrightarrow2x^2+6x+4-2x^2+7x-5=0\)
\(\Leftrightarrow13x-1=0\)
\(\Leftrightarrow13x=1\)
hay \(x=\frac{1}{13}\)
Vậy: \(x=\frac{1}{13}\)
