\(\dfrac{2}{x-1}=1+\dfrac{2x}{x+2}\Leftrightarrow\dfrac{2\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}+\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}\Leftrightarrow2x+4=x^2+x-2+2x^2-2x\)\(\Leftrightarrow2x-x^2-x-2x^2+2x=-2-4\Leftrightarrow3x-3x^2=-6\Leftrightarrow3x+3-3x^2+3=0\Leftrightarrow3\left(x+1\right)-3\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(3-3x+3\right)=0\Leftrightarrow\left(x+1\right)3\left(1-x\right)=0\Rightarrow\left[{}\begin{matrix}x+1=0\\1-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(\dfrac{2}{x-1}=1+\dfrac{2x}{x+2}\) ( ĐKXĐ: \(x\ne1\) , \(x\ne-2\) )
\(\Leftrightarrow\dfrac{2}{x-1}=\dfrac{x+2+2x}{x+2}\)
\(\Leftrightarrow\dfrac{2}{x-1}=\dfrac{3x+2}{x+2}\)
\(\Rightarrow2\left(x+2\right)=\left(3x+2\right).\left(x-1\right)\)
\(\Leftrightarrow2x+4=3x^2-3x+2x-2\)
\(\Leftrightarrow2x+4=3x^2-x-2\)
\(\Leftrightarrow3x^2-3x-6=0\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x^2+x\right)-\left(2x+2\right)=0\)
\(\Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)(thỏa mãn ĐKXĐ)
Vậy....
\(\dfrac{2}{x-1}=1+\dfrac{2x}{x+2}\)
\(\Rightarrow\dfrac{2}{x-1}=\dfrac{x+2+2x}{x+2}\)
\(\Rightarrow\dfrac{2}{x-1}=\dfrac{3x+2}{x+2}\)
\(\Rightarrow2.\left(x+2\right)=\left(x-1\right).\left(3x+2\right)\)
\(\Rightarrow2x+4=3x^2+2x-3x-1\)
\(\Rightarrow-3x^2+2x-2x-3x=-1-4\)
\(\Rightarrow-3x^2-3x=-5\)
\(\Rightarrow-3.\left(x^2+1\right)=-5\)
\(\Rightarrow x^2+1=\dfrac{5}{3}\Rightarrow x^2=\dfrac{2}{3}\)
\(\Rightarrow x=\pm\sqrt{\dfrac{2}{3}}\)
Vậy \(x=\pm\sqrt{\dfrac{2}{3}}\)
Chúc bạn học tốt!!! Mình cũng không chắc!!
\(\dfrac{2}{x-1}=1+\dfrac{2x}{x+2}\) (1)
ĐKXĐ : \(x\ne1;x\ne-2\)
\(\left(1\right)\Leftrightarrow\dfrac{2\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}\)
\(\Rightarrow2x+4=x^2+2x-x-2+2x^2-2x\)
\(\Leftrightarrow-3x^2+3x+6=0\)
\(\Leftrightarrow-3\left(x^2-x-6\right)=0\)
\(\Leftrightarrow3\left(x^2-3x+2x-6\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(KTMĐKXĐ\right)\\x=3\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vạy x=3.
= \(\dfrac{2}{x-1}-1-\dfrac{2x}{x+2}=0\)
= \(\dfrac{2\left(x+2\right)-\left(x-1\right)\left(x+2\right)-2x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=0\)
= \(\dfrac{2x+4-x^2-x+2-2x^2+2x}{\left(x-1\right)\left(x+2\right)}=0\)
= \(\dfrac{3x+6-3x^2}{\left(x-1\right)\left(x+2\right)}=0\)
= \(-3\left(x^2-x-2\right)=0\)
<=> \(x^2-x-2=0\)
Phương trình có dạng a \(-\)b + c =0
<=> 1 \(-\left(-1\right)\)+ \(\left(-2\right)\)= 0
Có 2 nghiệm X1= \(-1\) ; X2= 2
