1) ĐKXĐ: \(x\notin\left\{1;-2\right\}\)
Ta có: \(\dfrac{2x}{x-1}-\dfrac{1}{x+2}=2\)
\(\Leftrightarrow\dfrac{2x\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+2\right)}=\dfrac{2\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)
Suy ra: \(2x^2+4x-x+1=2\left(x^2+x-2\right)\)
\(\Leftrightarrow2x^2+3x+1=2x^2+2x-4\)
\(\Leftrightarrow2x^2+3x+1-2x^2-2x+4=0\)
\(\Leftrightarrow x+5=0\)
hay x=-5(thỏa ĐK)
Vậy: S={-5}
2) ĐKXĐ: \(x\notin\left\{5;-5\right\}\)
Ta có: \(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(x+x^2+5x-x-5=x-5\)
\(\Leftrightarrow x^2+5x-5-x+5=0\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
Vậy: S={0;-4}
a/ ĐKXĐ : \(x\ne1;-2\)
\(\dfrac{2x}{x-1}-\dfrac{1}{x+2}=2\)
\(\Leftrightarrow\dfrac{2x\left(x+2\right)-\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=2\)
\(\Leftrightarrow2x^2+3x-x+1=2x^2+4x-2x-4\)
\(\Leftrightarrow2x+1=2x-4\)
\(\Leftrightarrow1=-4\left(loại\right)\)
Vậy...
b/ĐKXĐ : \(x\ne\pm5\)
\(\dfrac{x}{x^2-25}-\dfrac{1-x}{x-5}=\dfrac{1}{x+5}\)
\(\Leftrightarrow\dfrac{x}{\left(x-5\right)\left(x+5\right)}+\dfrac{\left(x-1\right)\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow x+x^2+5x-x-5=x-5\)
\(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy...