Question

giải phương trình 

\(\dfrac{3}{x}+\dfrac{2x+1}{x+3}+\dfrac{x^2+12}{3x-x^2}=0\)

Answer 1

ĐKXĐ: \(x\notin\left\{0;3;-3\right\}\)

\(\dfrac{3}{x}+\dfrac{2x+1}{x+3}+\dfrac{x^2+12}{3x-x^2}=0\)

=>\(\dfrac{3\left(x+3\right)+x\left(2x+1\right)}{x\left(x+3\right)}-\dfrac{x^2+12}{x\left(x-3\right)}=0\)

=>\(\dfrac{2x^2+4x+9}{x\left(x+3\right)}-\dfrac{x^2+12}{x\left(x-3\right)}=0\)

=>\(\dfrac{\left(x-3\right)\left(2x^2+4x+9\right)-\left(x^2+12\right)\left(x+3\right)}{x\left(x+3\right)\left(x-3\right)}=0\)

=>\(2x^3+4x^2+9x-6x^2-12x-27-x^3-3x^2-12x-36=0\)

=>\(x^3-5x^2-15x-63=0\)

=>\(x\simeq7,91\left(nhận\right)\)