Question

Помогите пожалуйста
1 задание (b,d)

Answer 1

b)$\frac{x^{2}+4x }{x+2} = \frac{2x}{3} \\ 3( x^{2} +4x)=2x(x+2) \\ 3 x^{2} +12x=2 x^{2} +4x \\ 3 x^{2} -2 x^{2} +12x-4x=0 \\ x^{2} +8x=0 \\ x(x+8)=0 \\ x_{1}=0 \\ x_{2}+8=0 \\ x_{2}=-8$
d) $\frac{3}{ x^{2} +2} = \frac{1}{x}\\ 3x= x^{2} +2\\ x^{2} -3x+2=0 \\ \left \{ {{x_{1}*x_{2}=2} \atop x_{1}+x_{2}=3} \right. \\ x_{1}=2\\ x_{2}=1$