\(\dfrac{2}{x-1}=\dfrac{1+2x}{x+2}\)
\(\Leftrightarrow2\left(x+2\right)=\left(x-1\right)\left(2x+1\right)\)
\(\Leftrightarrow2x+4=2x^2-2x+x-1\)
\(\Leftrightarrow3x+4=2x^2-1\)
\(\Leftrightarrow2x^2-3x-4-1=0\)
\(\Leftrightarrow2x^2-3x-5=0\)
\(\Rightarrow x^2-\dfrac{3}{2}x-\dfrac{5}{2}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{49}{16}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right)^2-\dfrac{49}{16}=0\)
\(\Leftrightarrow\left(x-\dfrac{9}{16}\right)^2=\dfrac{49}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{7}{4}\\x-\dfrac{3}{4}=\dfrac{-7}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-1\end{matrix}\right.\)
Vậy...
Đúng 0
Bình luận (2)
