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Stepenovanje 5

TEKST ZADATKA

Dokazati identitete:

x−n(xn−1)−1+x−n(xn+1)−1=2(x2n−1)−1x^{-n}(x^n-1)^{-1} + x^{-n}(x^n+1)^{-1} = 2(x^{2n}-1)^{-1}
REŠENJE ZADATKA

Srediti izraz i i primeniti definiciju stepenovanja: a−m=1am, a^{-m}={\frac 1 {a^m}} , a=/ 0: a{=}\mathllap{/\,} 0 :

1xn1xn−1+1xn1xn+1=21x2n−1\frac 1 {x^n} \frac 1 {x^n-1} + \frac 1 {x^n} \frac 1 {x^n+1} = 2 \frac 1 {x^{2n}-1}

Srediti razlomak:

1xn(xn−1)+1xn(xn+1)=2x2n−1\frac 1 {x^n(x^n-1)} + \frac 1 {x^n(x^n+1)} = \frac 2 {x^{2n}-1}

Primeniti definiciju za sabiranje razlomaka | Primeniti definiciju za razliku kvadrata:

xn+1+xn−1xn(xn−1)(xn+1)=2(xn−1)(xn+1)\frac {x^n+1+x^n-1} {x^n(x^n-1)(x^n+1)} = \frac 2 {(x^n-1)(x^n+1)}

Srediti razlomak:

2xnxn(xn−1)(xn+1)=2(xn−1)(xn+1)\frac {2x^n} {x^n(x^n-1)(x^n+1)} = \frac 2 {(x^n-1)(x^n+1)}

Srediti razlomak:

2xnxn(xn−1)(xn+1)=2(xn−1)(xn+1)\frac {2\cancel{x^n}} {\cancel{x^n}(x^n-1)(x^n+1)} = \frac 2 {(x^n-1)(x^n+1)}

Identitet je dokazan:

2(xn−1)(xn+1)=2(xn−1)(xn+1),x=/ 0,x=/ ±1\frac 2 {(x^n-1)(x^n+1)} = \frac 2 {(x^n-1)(x^n+1)}, x {=}\mathllap{/\,} 0, x{=}\mathllap{/\,} \pm1

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