Các bạn giúp mk nha!
2x+1x2−2x+1 −2x+3x−1 =0
\(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}=0.\)
\(\frac{2x^2+3x+1}{\left(x-1\right)^2\left(x+1\right)}-\frac{2x^2-x+3}{\left(x-1\right)^2\left(x+1\right)}=0\)
\(\frac{2x+4}{\left(x-1\right)^2\left(x+1\right)}=0\)
=> 2x+4=0
2x=-4
x=-2
Học tốt nhé!
Dòng đầu mk cóp lại cái đầu bài nhưng nó bị lỗi đấy
\(\frac{2x+1}{x^2-2x+1}-\frac{2x+3}{x-1}=0ĐK:x\ne\pm1\)
\(\Leftrightarrow\frac{2x+1\left(x-1\right)}{\left(x-1\right)^2\left(x-1\right)}-\frac{\left(2x+3\right)\left(x-1\right)^2}{\left(x-1\right)\left(x-1\right)^2}=0\)
Khử mẫu : \(\left(2x+1\right)\left(x-1\right)-\left(2x+3\right)\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1-\left(2x+3\right)\left(x-1\right)\right)=0\)
TH1 : \(x-1=0\Leftrightarrow x=1\)
TH2 : \(2x+1-\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow x+4-2x^2=0\) ( giải delta )
\(\Delta=1^2-4.4.\left(-2\right)=1+32=33>0\)
\(x_1=\frac{-1-\sqrt{33}}{-2};x_2=\frac{-1+\sqrt{33}}{-2}\)
