\(\frac{x+1}{x-3}-\frac{1}{x-1}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow x^2-1-x+3-2=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
\(\frac{x+1}{x-3}-\frac{1}{x-1}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
điều kiện: \(x\ne3;1\)
quy đồng mẫu hai phân số
\(\frac{\left(x+1\right)\left(x-1\right)-\left(x-3\right)}{\left(x-3\right)\left(x-1\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\frac{x^2-1-x+3}{\left(x-3\right)\left(x-1\right)}-\frac{2}{\left(x-1\right)\left(x-3\right)}=0\)
\(\frac{x^2-1-x+3-2}{\left(x-1\right)\left(x-3\right)}=0\)
\(\frac{x^2-x}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Rightarrow x^2-x=0\)
\(x\left(x-1\right)=0\)
vây x = 0
hoặc x = 1 (không thỏa điều kiện)
vậy x = 0
\(ĐKXĐ:x\ne1;x\ne3\)
\(\frac{x+1}{x-3}-\frac{1}{x-1}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\Rightarrow\frac{\left(x+1\right)\left(x-1\right)-\left(x-3\right)}{\left(x-3\right)\left(x-1\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\Rightarrow\frac{x^2-1-x+3}{\left(x-3\right)\left(x-1\right)}=\frac{2}{\left(x-1\right)\left(x-3\right)}\)
\(\Rightarrow x^2-1-x+3=2\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\in\left\{0;1\right\}\)
\(\frac{x+1}{x-3}-\frac{1}{x-1}-\frac{2}{\left(x-1\right)\left(x-3\right)}=0\left(x\ne1;x\ne3\right)\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\frac{x-3}{\left(x-1\right)\left(x-3\right)}-\frac{2}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2-1-x+3-2}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2-x}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Rightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=1\left(ktm\right)\end{cases}}}\)
Vậy x=0
