Question

Efectuaţi pe DVA:
d.)-125 supra x^2-5x + x^2 supra x-5
f.)(1 supra x-1) totul la puterea -1 + x supra x+1 - x+2 supra x^2-1

Va rog frumos ajutatima

Answer 1

d) $\frac{-125}{ x^{2} -5x} + \frac{ x^{2} }{x-5}= \frac{-125}{ x(x -5)} + \frac{ x^{2} }{x-5}$ amplificam a doua fractie cu x ==>

Answer 2

   
[tex]\displaystyle d) \frac{-125}{x^2-5}+ \frac{x^2}{x-5} = ~~~(unde~~ x \neq 0~si ~ x \neq 5) \\ \\ = \frac{-125}{x(x-5)}+ \frac{x^2}{x-5} = = \frac{-125}{x(x-5)}+ \frac{x \cdot x^2}{x(x-5)} = \frac{x^3-125}{x(x-5)}= \\ \\ = \frac{(x-5)(x^2+5x+25) }{x(x-5)}= \boxed{\frac{x^2+5x+25 }{x}}[/tex]


[tex]\displaystyle f) \\ \left(\frac{1}{x-1} \right)^{-1}+ \frac{x}{x+1} - \frac{x+2}{x^2 - 1} = \\ \\ =\frac{x-1}{1}+ \frac{x}{x+1} - \frac{x+2}{x^2 - 1} = \\ \\ =\frac{(x-1)(x^2 - 1)}{x^2 - 1}+ \frac{x(x-1)}{x^2-1} - \frac{x+2}{x^2 - 1} = \\ \\ =\frac{(x-1)(x^2 - 1) + x(x-1) -(x+2)}{x^2 - 1} =\\ \\ =\frac{x^3-x^2 -x + 1 + x^2 -x -x-2}{x^2 - 1} = \\ \\ = \boxed{ \frac{x^3 -3x-1 }{x^2 - 1} }[/tex]