Question

пжпжпжппжпж 100 балоов
спам бан, с росписью действий

Answer 1

вот это правда друг 100%

Answer 2

$\displaystyle\bf\\1)\\\\\frac{x^{2}-6x-7 }{x-7} =0\\\\\\\left \{ {{x^{2} -6x-7=0} \atop {x-7\neq 0}} \right. \\\\\\\left \{ {{((x+1)(x-7)=0} \atop {x\neq 7}} \right. \\\\\\\left \{ {{\left[\begin{array}{ccc}x_{1} =-1\\x_{2} =7-neyd\end{array}\right } \atop {x\neq 7}} \right. \\\\\\Otvet \ : \ -1\\\\2)\\\\\frac{x^{2} +10x}{x-8} =\frac{12x+48}{x-8} \\\\\\\left \{ {{x^{2} +10x=12x+48} \atop {x-8\neq 0}} \right. \\\\\\\left \{ {{x^{2} -2x-48=0} \atop {x\neq 8}} \right.$

$\displaystyle\bf\\\left \{ {{(x+6)(x-8)=0} \atop {x\neq 8}} \right. \\\\\\\left \{ {{\left[\begin{array}{ccc}x_{1} =-6\\x_{2} =8-neyd\end{array}\right } \atop {x\neq 8}} \right. \\\\\\Otvet \ : \ -6\\\\3)\\\\\frac{3x+4}{x-3} =\frac{2x-9}{x+1} \\\\\\\frac{3x+4}{x-3} -\frac{2x-9}{x+1} =0\\\\\\\frac{(3x+4)\cdot(x+1)-(2x-9)\cdot(x-3)}{(x-3)(x+1)} =0\\\\\\\frac{3x^{2} +3x+4x+4-2x^{2} +6x+9x-27}{(x-3)(x+1)} =0\\\\\\\frac{x^{2} +22x-23}{(x-3)(x+1)}=0$

$\displaystyle\bf\\\left \{ {{x^{2} +22x-23=0} \atop {x-3\neq 0 \ ; \ x+1\neq 0}} \right. \\\\\\\left \{ {{(x-1)\cdot(x+23)=0} \atop {x\neq 3 \ ; \ x\neq -1}} \right. \\\\\\\left \{ {{\left[\begin{array}{ccc}x_{1} =1\\x_{2} =-23\end{array}\right} \atop {x\neq 3 \ ; \ x\neq -1}} \right. \\\\\\Otvet \ : \ 1 \ ; \ -23$