Question

tìm x biết 2x+1/(x-1)^2  -   2x+3/x^2-1 = 0

Answer 1

Điều kiện: \(x\ne\pm1\)

\(\frac{2x+1}{\left(x-1\right)^2}-\frac{2x+3}{x^2-1}=0\)

\(\Rightarrow\frac{\left(2x+1\right)\left(x+1\right)-\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}=0\)

\(\Rightarrow\frac{2x^2+3x+1-\left(2x^2+x-3\right)}{\left(x-1\right)^2\left(x+1\right)}=0\)

\(\Rightarrow\frac{2x+4}{\left(x-1\right)^2\left(x+1\right)}=0\Rightarrow2x+4=0\Rightarrow x=-2\)(thỏa mãn điều kiện)

Vậy x = -2