ЗПЗ §§ 5–8 Алгебра = ЗПЗ §§ 10–13 Математика
Розв’яжіть рівняння:
1. $\frac{4x+8}{x-3}=0;$
2. $\frac{4x^2-8}{x+1}=4x;$
Розв'язок:
1. $\frac{4x+8}{x-3}=0;\ \left\{\begin{matrix}\left.4x+8=0,\right.\\\left.x-3\neq0;\right.\\\end{matrix}\right.$
$ \left\{\begin{matrix}\left.4x=-8,\right.\\\left.x\neq3;\right.\\\end{matrix}\right.\ \ \left\{\begin{matrix}\left.x=-2,\right.\\\left.x\neq3;\right.\\\end{matrix}\right.\ \ $
2. $\frac{4x^2-8}{x+1}=4x;$
$\frac{4x^2-8}{x+1}=\frac{4x}{1}; x≠-1; $
$\left\{\begin{matrix}\left.4x^2-8=4x(x+1),\right.\\\left.x+1\neq0;\right.\\\end{matrix}\right.$
$\left\{\begin{matrix}\left.4x^2-8=4x^2+4x,\right.\\\left.x\neq-1;\right.\\\end{matrix}\right. $
$\left\{\begin{matrix}\left.4x^2-4x^2-4x=8,\right.\\\left.x\neq-1;\right.\\\end{matrix}\right.$
$\left\{\begin{matrix}\left.-4x=8,\right.\\\left.x\neq-1;\right.\\\end{matrix}\right.\ \ $
$\left\{\begin{matrix}\left.x=-2,\right.\\\left.x\neq-1;\right.\\\end{matrix}\right.\ \ x=-2.$
Відповідь:
1. –2;
2. –2.