Giải các phương trình sau: b. x+3/x+1 - 3x^2 + 4x +1/ x(x+1) = x-1/x

b) ĐKXĐ:$\left\{\begin{array}{c}x\ne 0\\ x+1\ne 0\end{array}\right.\iff \left\{\begin{array}{c}x\ne 0\\ x\ne -1\end{array}\right.$

Khi đó phương trình đã cho tương đương với:

$\frac{\left(x+3\right)x}{x\left(x+1\right)}-\frac{3x^2+4x+1}{x\left(x+1\right)}=\frac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}$

$\iff \frac{x^2+3x}{x\left(x+1\right)}-\frac{3x^2+4x+1}{x\left(x+1\right)}=\frac{x^2-1}{x\left(x+1\right)}$

$\iff \frac{x^2+3x-3x^2-4x-1}{x\left(x+1\right)}=\frac{x^2-1}{x\left(x+1\right)}$

$\iff \frac{-2x^2-x-1}{x\left(x+1\right)}=\frac{x^2-1}{x\left(x+1\right)}$

Þ -2x² - x - 1 = x² - 1

Û 3x² + x = 0

Û x(3x + 1) = 0

$\Rightarrow \left[\begin{array}{c}x=0\\ 3x+1=0\end{array}\right.\iff \left[\begin{array}{c}x=0\left(L\right)\\ x=\frac{-1}{3}\left(TMÐK\right)\end{array}\right.$