Question

Решите уравнение: б) 3/х^2+4х-15/х2-4х=4/х

Answer 1

$\frac{3}{x^{2} +4x}-\frac{15}{x^{2} -4x}=\frac{4}{x}\\\\\frac{3}{x(x+4)}-\frac{15}{x(x-4)}-\frac{4}{x}=0\\\\\frac{3x-12-15x-60-4x^{2}+64 }{x(x-4)(x+4)}=0\\\\\frac{4x^{2}+12x+8 }{x(x-4)(x+4)}=0\\\\\frac{x^{2}+3x-2 }{x(x-4)(x+4)}=0\\\\\left \{ {{x^{2}+3x+2=0 } \atop {x\neq0;x\neq-4;x\neq4}} \right.\\\\D=(-3)^{2}-4*(2)=9-8=1\\\\x_{1}=\frac{-3+1 }{2}=-1\\\\x_{2}=\frac{-3-1 }{2}=-2$