Question

Помогите:2,3,4 задание!?)
Заранее спасибо

Answer 1

№2
$a) \frac{x^2}{x^2-1}- \frac{x}{x+1}= \frac{x^2}{(x-1)(x+1)} - \frac{x}{x+1}=$
$\frac{x^2}{(x-1)(x+1)}- \frac{x}{x+1}= \frac{x^2}{(x-1)(x+1)}- \frac{x(x-1)}{(x-1)(x+1)} = \frac{x^2-x(x-1)}{(x-1)(x+1)} =$
$\frac{x^2-x^2+x}{(x-1)(x+1)}= \frac{x}{(x-1)(x+1)}$
$b) \frac{c^2}{c^2-4}- \frac{c}{c-2} = \frac{c^2}{(c-2)(c+2)} - \frac{c}{c-2}= \frac{c^2}{(c-2)(c+2)}- \frac{c(c+2)}{(c-2)(c+2)} =$
$\frac{c^2-c(c+2)}{(c-2)(c+2)}= \frac{c^2-c^2-2c}{(c-2)(c+2)}= \frac{2c}{(2-c)(c+2)}$

№3
$a)10x^2+5x=0\\5x(2x+1)=0\\5x=0.......ore....2x+1=0\\x=0.......ore.....x=-0.5\\OTBeT:-0.5;0.$
$b)12x^2+3x=0\\3x(4x+1)=0\\4x+1=0.....ore.....3x=0\\4x=-1.....ore.....x=0\\x=-0.25.....ore.....x=0\\OTBeT:-0.25;0.$
№4
$\left \{ {{2- \frac{3+2x}{3}\ \textgreater \ 1- \frac{x+6}{2}|*6 } \atop {3- \frac{x}{4} \ \textless \ x|*4}} \right. \left \{ {{12-6-4x\ \textgreater \ 6-3x-18} \atop {12-x\ \textless \ 4x}} \right. \left \{ {{-x\ \textgreater \ -18} \atop {-5x\ \textless \ -12}} \right. \left \{ {{x\ \textless \ 18} \atop {x\ \textgreater \ 2.4}} \right. \left \{ {+++++(18)\ \textgreater \ x} \atop {-(2.4)++++\ \textgreater \ x}} \right.$
$x$∈$(2.4;18)$