Question

$\frac{24}{x^{2} + 4x} - \frac{15}{x^{2} - 4x } = \frac{11}{x}$
помогите!!!!​

Answer 1

Ответ:

Смотри решение на фото

Answer 2

$\frac{24}{x^{2}+4x }-\frac{15}{x^{2}-4x } =\frac{11}{x}\\\\\frac{24}{x(x+4)}-\frac{15}{x(x-4)}-\frac{11}{x} =0\\\\\frac{24(x-4)-15(x+4)-11(x^{2}-16)}{x(x+4)(x-4)}=0\\\\\frac{24x-96-15x-60-11x^{2}+176 }{x(x+4)(x-4)}=0\\\\\frac{11x^{2}+9x+20 }{x(x+4)(x-4)} =0\\\\x\neq 0 \ ; \ x\neq-4 \ ; \ x\neq4\\\\11x^{2}-9x-20=0\\\\D=9^{2}-4*11*(-20)=81+880=961=31^{2} \\\\x_{1}=\frac{9+31}{22} =1\frac{9}{11}\\\\x_{2}=\frac{9-31}{22}=-1$