Предмет: Алгебра, автор: eminem90

Решите пожалуйста номер 3 все примеры дам 60 баллов!!!!!

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Ответы

Автор ответа: Universalka

$\frac{x^{2}-16 }{x+4}=0 \\\\\left \{ {{x^{2}-16=0 } \atop {x+4\neq0 }} \right.\\\\\left \{ {{(x-4)(x+4)=0} \atop {x\neq-4 }} \right.\\\\\left \{ {{\left[\begin{array}{ccc}x_{1}=4 \\x_{2} =-4-neyd\end{array}\right } \atop {x\neq-4 }} \right.\\\\Otvet:\boxed4\\\\\\2)\frac{5}{x+7}-\frac{3}{x-7}=0\\\\\frac{5*(x-7)-3*(x+7)}{(x+7)(x-7)}=0\\\\\frac{5x-35-3x-21}{(x+7)(x-7)}=0\\\\\frac{2x-56}{(x+7)(x-7)}=0\\\\\left \{ {{2x-56=0} \atop {x\neq-7;x\neq 7 }} \right.\\\\x-28=0\\\\x=28$

$Otvet:\boxed{28}\\\\\\3)\frac{x-2}{x+3}=\frac{4x-1}{4x+1}\\\\\frac{x-2}{x+3}-\frac{4x-1}{4x+1}=0\\\\\frac{(x-2)(4x+1)-(4x-1)(x+3)}{(x+3)(4+1)}=0\\\\\frac{4x^{2} +x-8x-2-4x^{2}-12x+x+3 }{(x+3)(4x+1)} =0\\\\\frac{-18x+1}{(x+3)(4x+1)}=0\\\\\left \{ {{-18x+1=0} \atop {x\neq-3;x\neq-0,25}} \right.\\\\-18x=-1\\\\x=\frac{1}{18}\\\\Otvet:\boxed{\frac{1}{18}}$

$4)\frac{4x-3}{x+1}-\frac{6x-5}{2x+1}=1\\\\\frac{4x-3}{x+1}-\frac{6x-5}{2x+1}-1=0\\\\\frac{(4x-3)(2x+1)-(6x-5)(x+1)-(x+1)(2x+1)}{(x+1)(2x+1)}=0\\\\\frac{8x^{2}+4x-6x-3-6x^{2}-6x+5x+5-2x^{2}-x-2x-1}{(x+1)(2x+1)}=0\\\\\frac{-6x+1}{(x+1)(2x+1)}=0\\\\\left \{ {{-6x+1=0} \atop {x\neq-1;x\neq-0,5 }} \right.\\\\-6x+1=0\\\\-6x=-1\\\\x=\frac{1}{6} \\\\Otvet:\boxed{\frac{1}{6}}$

$5)\frac{1}{x^{2}-6x }+\frac{1}{x^{2}+6x }=\frac{2x}{x^{2}-36}\\\\\frac{1}{x(x-6)}+\frac{1}{x(x+6)}-\frac{2x}{(x-6)(x+6)}=0\\\\\frac{x+6+x-6-2x^{2}}{(x-6)(x+6)}=0\\\\\frac{2x-2x^{2}}{(x-6)(x+6)}=0\\\\\frac{x-x^{2}}{(x-6)(x+6)}=0\\\\\frac{x(1-x)}{(x-6)(x+6)}=0\\\\\left \{ {{x_{1}=0;x_{2}=1 } \atop {x\neq-6;x\neq6 }} \right.$