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Forkorte...

Posted: 29/08-2015 09:48
by iBrus
Hvilke steg gjøres for å forkorte denne:
[tex]z = \frac{2}{5}(x+1)^{\frac{5}{2}} - \frac{2}{3}(x+1)^{\frac{3}{2}} + C[/tex]

til denne?:
[tex]z = \frac{2}{15}(x+1)^{\frac{3}{2}}(3x-2) + C[/tex]


Ser ikke helt hvordan...


iBrus

Re: Forkorte...

Posted: 29/08-2015 10:56
by Fibonacci92
Hint: $(x+1)^{\frac{5}{2}} = (x+1)^{\frac{3}{2}}\cdot (x+1)$

Re: Forkorte...

Posted: 29/08-2015 11:00
by DennisChristensen
iBrus wrote:Hvilke steg gjøres for å forkorte denne:
[tex]z = \frac{2}{5}(x+1)^{\frac{5}{2}} - \frac{2}{3}(x+1)^{\frac{3}{2}} + C[/tex]

til denne?:
[tex]z = \frac{2}{15}(x+1)^{\frac{3}{2}}(3x-2) + C[/tex]


Ser ikke helt hvordan...


iBrus
$ z = \frac{2}{5}(x+1)^{\frac{5}{2}} - \frac{2}{3}(x+1)^{\frac{3}{2}} + C \\
= (x+1)^{\frac{3}{2}}\left(\frac{2}{5}(x+1) - \frac{2}{3}\right) + C \\
= (x+1)^{\frac{3}{2}}\left(\frac{6}{15}(x+1) - \frac{10}{15}\right) + C \\
= \frac{2}{15}(x+1)^{\frac{3}{2}}\left(3(x+1) - 5\right) + C \\
=\frac{2}{15}(x+1)^{\frac{3}{2}}(3x-2) + C$

Re: Forkorte...

Posted: 29/08-2015 12:00
by iBrus
Takker...

iBrus