Решите Уравнение Пожалуйста[latex]a)\frac{x^{2}}{13+x} = \frac{3x}{13+x}   2) \frac{ x^{2}-1 }{x+5} = \frac{5-x}{x+5}     3) \frac{4 x^{2} -11x-3}{3-x} =0 [/latex]

[latex]1).....\; \; \left \{ {{x^2=3x} \atop {x+13\ne 0}} \right. \; \left \{ {{x(x-3)=0} \atop {x\ne -13}} \right. \; \Rightarrow \; x_1=0,\; x_2=3\\\\2)\; ... \left \{ {{x^2-1=5-x} \atop {x+5\ne 0}} \right. \; \left \{ {{x^2-x-6=0} \atop {x\ne -5}} \right. \; \left \{ {{x_1=-2,x_2=3\; (t.\; Vieta)} \atop {x\ne -5}} \right. \; \Rightarrow \; x_1=-2,x_2=3[/latex] [latex]3)\; .....\; \left \{ {{4x^2-11x-3=0} \atop {3-x\ne 0}} \right. \; \left \{ {{x_1=-\frac{1}{4},x_2=3} \atop {x\ne 3}} \right. \; \Rightarrow \; \; x=-\frac{1}{4}\\\\D=11^2+4\cdot 4\cdot 3=169,\\\\x_1=\frac{11-13}{8}=\frac{-2}{8}=-\frac{1}{4},\; \; x_2=\frac{11+13}{8}=3[/latex]