Question

ma poate ajuta cineva? va rog e urgent.. nu pot sa imi dau seama cum da rezultatul

Answer 1

$\displaystyle a)E(x)=\left( \frac{x+2}{x^2-x} + \frac{x-3}{x^2-1} - \frac{x-1}{x^2+x} \right): \frac{x^2+3x+2}{x^3-2x^2+x} \\ \\ E(x)=\left( \frac{x+2}{x(x-1)} + \frac{x-3}{(x-1)(x+1)} - \frac{x-1}{x(x+1)} \right): \frac{(x+1)(x+2)}{x(x^2-2x+1)} \\ \\ E(x)= \frac{(x+2)(x+1)+x(x-3)-(x-1)(x-1)}{x(x-1)(x+1)} \cdot \frac{x(x-1)^2}{(x+1)(x+2)}$
$\displaystyle E(x)= \frac{x^2+x+2x+2+x^2-3x-x^2+2x-1}{x+1} \cdot \frac{x-1}{(x+1)(x+2)} \\ \\ E(x)= \frac{x^2+2x+1}{x+1} \cdot \frac{x-1}{(x+1)(x+2)} \\ \\ E(x)= \frac{(x+1)^2}{x+1} \cdot \frac{x-1}{(x+1)(x+2)} \\ \\ E(x)= \frac{x-1}{x+2}$
$\displaystyle b)E(x)= \frac{3(x+1)}{x+2} \Rightarrow \frac{x-1}{x+2} =\frac{3(x+1)}{x+2} \Rightarrow \frac{x-1}{x+2} = \frac{3x+3}{x+2} \Rightarrow \\ \\ \Rightarrow (x-1)(x+2)=(x+2)(3x+3) \Rightarrow \\ \\ \Rightarrow x^2+2x-x-2=3x^2+3x+6x+6 \Rightarrow \\ \\ \Rightarrow x^2+2x-x-2-3x^2-3x-6x-6=0 \Rightarrow -2x^2-8x-8=0|\cdot(-1) \Rightarrow\\ \\ \Rightarrow 2x^2+8x+8=0|:(-2)\Rightarrow x^2+4x+4=0 \\ \\a=1,~b=4,~c=4\\ \\ \Delta=b^2-4ac=4^2-4\cdot 1 \cdot4=16-16=0\\ \\x_1=x_2= \frac{-b}{2a} = \frac{-4}{2\cdot1}=-2$
$\displaystyle x \in R-\{-2\} \Rightarrow x \in \varnothing$