\(\frac{x+2}{2x+1}=\frac{x}{2x-1}\) (ĐK:\(x\ne\pm\frac{1}{2}\))
\(\Leftrightarrow\)\(\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\frac{x\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(\Leftrightarrow\)\(\left(x+2\right)\left(2x-1\right)=x\left(2x+1\right)\)
\(\Leftrightarrow2x^2-x+4x-2=2x^2+x\)
\(\Leftrightarrow2x^2-x+4x-2-2x^2-x=0\)
\(\Leftrightarrow2x-2=0\)
\(\Leftrightarrow2\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\left(TM\right)\)
Vậy x=1
\(\frac{x+2}{2x+1}=\frac{x}{2x-1}\)
Ta có:(x+2)(2x-1)=x.(2x+1)
2x2+3x-2=2x2+x
Sau đó tự làm
\(\frac{x+2}{2x+1}=\frac{x}{2x-1}\)
điều kiện \(x\ne\pm\frac{1}{2}\)
<=>(x+2)(2x-1)=x(2x+1)
<=>2x2+3x-2=2x2+x
<=> 2x-2=0<=>x=1
\(\frac{x+2}{2x+1}=\frac{x}{2x-1}\)
\(ĐKXĐ:x\ne\pm\frac{-1}{2}\)
\(\Leftrightarrow\frac{x+2}{2x+1}-\frac{x}{2x-1}=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}-\frac{x\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}=0\)
\(\Rightarrow2x^2-x+4x-2-2x^2-x=0\)
\(\Leftrightarrow2x-2=0\)
\(\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\left(tm\right)\)
\(\text{Vậy x=1}\)
