Question

Помогите пожалуйста 8 клас

Answer 1

$\displaystyle\bf\\1)\\\\\frac{x^{2} +5x}{x-2} =0\\\\\\\frac{x\cdot(x+5)}{x-2} =0\\\\\\\left \{ {{x\cdot(x+5)=0} \atop {x-2\neq 0}} \right. \\\\\\\left\{\begin{array}{ccc}x=0\\x+5=0\\x\neq 2\end{array}\right\\\\\\\left\{\begin{array}{ccc}x_{1} =0\\x_{2}=-5 \\x\neq 2\end{array}\right\\\\\\Otvet \ : \ 0 \ ; \ -5$

$\displaystyle\bf\\2)\\\\\frac{x}{x+1} =\frac{2x+3}{2x-5} \\\\\\x\cdot(2x-5)=(x+1)\cdot(2x+3) \ \ \ , \ \ x\neq -1 \ \ ; \ \ x\neq 2,5\\\\\\2x^{2}-5x=2x^{2} +3x+2x+3 \\\\\\2x^{2}-5x-2x^{2} -5x=3 \\\\\\0\cdot x=3\\\\\\Otvet \ : \ kornei \ net$