\(\dfrac{2}{x-1}=1+\dfrac{2x}{x+2}\)
\(ĐK:x\ne1;-2\)
\(\Leftrightarrow\dfrac{2\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x+2\right)+2x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}\)
\(\Leftrightarrow2\left(x+2\right)=\left(x-1\right)\left(x+2\right)+2x\left(x-1\right)\)
\(\Leftrightarrow2x+4=x^2+2x-x-1+2x^2-2x\)
\(\Leftrightarrow3x^2-3x-5=0\)
Nghiệm xấu lắm bạn!
Ta có \(\dfrac{2}{x-1}=1+\dfrac{2x}{x+2}\)
<=> \(\dfrac{2\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}\)
=>\(2x+4=x^2+2x-x-2+2x^2-2x\)
<=>\(2x-x^2-2x+x-2x^2+2x=-2-4\)
<=>\(-3x^2+3x=-6\)
<=> \(-3x^2+3x+6=0\)
<=>- \(3\left(x^2-x-6\right)=0\)
<=> \(x^2+2x-3x-6=0\)
<=>\(x\left(x+2\right)-3\left(x+2\right)\)
<=> \(\left(x+2\right)\left(x-3\right)=0\)
<=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Vậy tập nghiệm của pt là S={-2;3}
Chúc bạn học tốt <3
