Question

$Aratati~ca~: \\ \\ \frac{1-a}{1-a^{-1}} + \frac{1+a^{-1}}{1-a} + \frac{1+a}{1-a^{-1}} + \frac{1-a^{-1}}{1+a} =4$

Answer 1

   
[tex]\displaystyle\\ \text{Prima fractie are la numitor: }~(1+a^{-1})~~\text{in loc de } (1-a^{-1})\\\\\frac{1-a}{1+a^{-1}} + \frac{1+a^{-1}}{1-a} + \frac{1+a}{1-a^{-1}} + \frac{1-a^{-1}}{1+a} = \\ \\ \text{Grupam prima fractie cu a treia si a doua cu a patra.}\\\\ =\frac{1-a}{1+a^{-1}} + \frac{1+a}{1-a^{-1}} + \frac{1+a^{-1}}{1-a} + \frac{1-a^{-1}}{1+a} =\\\\ \text{Eliminam exponentul negativ.} [/tex]


[tex]\displaystyle\\ =\frac{1-a}{1+ \frac{1}{a}} + \frac{1+a}{1-\frac{1}{a}} + \frac{1+\frac{1}{a}}{1-a} + \frac{1-\frac{1}{a}}{1+a} =\\\\ =\frac{1-a}{\frac{a+1}{a}} + \frac{1+a}{\frac{a-1}{a}} + \frac{\frac{a+1}{a}}{1-a} + \frac{\frac{a-1}{a}}{1+a} =\\\\ =\frac{a(1-a)}{a+1} + \frac{a(1+a)}{a-1} + \frac{a+1}{a(1-a)} + \frac{a-1}{a(1+a)} =\\\\ \text{Inversam expresiile de forma (1 - a) si (1 + a) in (a - 1) si (a + 1).}\\\\ =-\frac{a(a-1)}{a+1} + \frac{a(a+1)}{a-1} - \frac{a+1}{a(a-1)} + \frac{a-1}{a(a+1)} =[/tex]


[tex]\displaystyle\\ -\frac{a(a-1)}{a+1} + \frac{a(a+1)}{a-1} - \frac{a+1}{a(a-1)} + \frac{a-1}{a(a+1)} =\\\\ \text{Numitorul comun al fractiilor este: } ~a(a^2-1)\\\\ -\frac{^{^{a(a-1)}}^)a(a-1)}{a+1} + \frac{^{^{a(a+1)}}^)a(a+1)}{a-1} - \frac{^{^{(a+1)}}^)a+1}{a(a-1)} + \frac{^{^{(a-1)}}^)a-1}{a(a+1)} =\\\\ -\frac{a^2(a-1)^2}{a(a^2-1)} + \frac{a^2(a+1)^2}{a(a^2-1)} - \frac{(a+1)^2}{a(a^2-1)} + \frac{(a-1)^2}{a(a^2-1)} =\\\\ =\frac{-a^2(a-1)^2+a^2(a+1)^2-(a+1)^2+(a-1)^2}{a(a^2-1) } = [/tex]


[tex]\displaystyle\\\ =\frac{-a^2(a-1)^2+a^2(a+1)^2-(a+1)^2+(a-1)^2}{a(a^2-1) } =\\\ =\frac{-a^2(a^2-2a+1)+a^2(a^2+2a+1)-(a^2+2a+1)+a^2-2a+1}{a(a^2-1) }=\\\\ =\frac{-(a^4-2a^3+a^2)+a^4+2a^3+a^2-(a^2+2a+1)+a^2-2a+1}{a(a^2-1) }=\\\\ =\frac{-a^4 +2a^3-a^2+a^4+2a^3+a^2-a^2-2a-1+a^2-2a+1}{a(a^2-1) }=\\\\ \text{La numarator se reduc termenii: }~~ \pm a^4;~\pm a^2;~\pm1 \\\\ =\frac{2a^3+2a^3-2a-2a}{a(a^2-1) }=\frac{4a^3-4a}{a(a^2-1) }=4\frac{a^3-a}{a(a^2-1) }=4\frac{a(a^2-1)}{a(a^2-1) }=\boxed{\texttt{4}}[/tex]

Answer 2

Înlocuim a⁻¹ = 1/a și fiecare fracție devine:

$\it I)\ \ \dfrac{1-a}{1+\dfrac{1}{a}} = \dfrac{1-a}{\dfrac{a+1}{a}} =\dfrac{a(1-a)}{a+1}$


$\it II)\ \ \dfrac{1+\dfrac{1}{a}}{1-a} = \dfrac{\dfrac{a+1}{a}}{1-a} =\dfrac{a+1}{a(1-a)} =\dfrac{-(a+1)}{a(a-1)}$



$\it III)\ \ \dfrac{1+a}{1-\dfrac{1}{a}} = \dfrac{1+a}{\dfrac{a-1}{a}} =\dfrac{a(a+1)}{a-1}$



$\it \ \ \dfrac{1-\dfrac{1}{a}}{1+a} =\dfrac{\dfrac{a-1}{a}}{a+1} =\dfrac{a-1}{a(a+1)}$

Adunăm prima cu a patra fracție :


[tex]\it \dfrac{a(1-a)}{a+1} +\dfrac{a-1}{a(a+1)} =\dfrac{a^2(1-a)+(a-1)}{a(a+1)} =\dfrac{a^2(1-a) -(1-a)}{a(a+1)} = \\\;\\ \\\;\\ =\dfrac{(1-a)(a^2-1)}{a(a+1)} = \dfrac{(1-a)(a-1)(a+1)}{a(a+1)} = \dfrac{-(a-1)^2}{a}[/tex]

Adunăm a doua cu a treia fracție:


[tex]\it \dfrac{-(a+1)}{a(a-1)}+\dfrac{a(a+1)}{a-1} = \dfrac{-(a+1)+a^2(a+1)}{a(a-1)} =\dfrac{(a+1)(a^2-1)}{a(a-1)}= \\\;\\ \\\;\\ = \dfrac{(a+1)(a-1)(a+1)}{a(a-1)} =\dfrac{(a+1)^2}{a}[/tex]

Acum, expresia devine:


[tex]\it \dfrac{(a+1)^2}{a} +\dfrac{-(a-1)^2}{a} = \dfrac{(a+1)^2 - (a-1)^2}{a} = \\\;\\ \\\;\\ = \dfrac{a^2+2a+1-a^2+2a-1}{a} =\dfrac{4a}{a} =4[/tex]