1)
\(ĐK:x\ne1;x\ne0\)
PT trở thành:
\(\dfrac{x.x}{x\left(x-1\right)}+\dfrac{\left(x-1\right)\left(x-1\right)}{x\left(x-1\right)}=2\\ \Leftrightarrow\dfrac{x^2}{x\left(x-1\right)}+\dfrac{\left(x-1\right)^2}{x\left(x-1\right)}=\dfrac{2x\left(x-1\right)}{x\left(x-1\right)}\\ \Leftrightarrow x^2+x^2-2x+1-2x^2+2x=0\\\Leftrightarrow1=0\left(vô.lý\right)\)
Vậy PT vô nghiệm
3)
ĐK: \(x\ne\pm1\)
PT trở thành:
\(\dfrac{2\left(x-1\right)}{x^2-1}-\dfrac{3\left(x+1\right)}{x^2-1}=\dfrac{5\left(x^2-1\right)}{x^2-1}\\ \Leftrightarrow2x-2-3x-3-5x^2+5=0\\ \Leftrightarrow-5x^2-x=0\\ \Leftrightarrow-x\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-\dfrac{1}{5}\left(nhận\right)\end{matrix}\right.\)
13:
\(\dfrac{x-1}{x}-\dfrac{3x}{2x-2}=\dfrac{5}{2}\)
=>\(\dfrac{2\left(x-1\right)^2-3x^2}{2x\left(x-1\right)}=\dfrac{5x\left(x-1\right)}{2x\left(x-1\right)}\)
=>5x^2-5x=2x^2-4x+2-3x^2
=>5x^2-5x=-x^2-4x+2
=>6x^2-x-2=0
=>6x^2-4x+3x-2=0
=>(3x-2)(2x+1)=0
=>x=2/3 hoặc x=-1/2
11:
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
=>\(\dfrac{x\left(x+2\right)-x+2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
=>x^2+2x-x+2=2
=>x^2+x=0
=>x(x+1)=0
=>x=0(loại) hoặc x=-1(nhận)
