Question

помогите пжпжпжпжпжпжппжпжпж​

Answer 1

$\displaystyle\bf\\x^{2} +\frac{1-3x}{x+4} =16-\frac{3x-1}{x+4} \\\\\\x^{2} +\frac{1-3x}{x+4} =16+\frac{1-3x}{x+4} \\\\\\x^{2} +\frac{1-3x}{x+4} -\frac{1-3x}{x+4}=16\\\\\\x^{2} =16 \ \ \ ; \ \ x+4\neq 0 \ \ \ \Rightarrow \ \ \ x\neq -4\\\\\\x_{1,2} =\pm \ \sqrt{16} =\pm \ 4\\\\x_{1} =4\\\\x_{2} =-4-neyd\\\\Otvet \ : \ 4$