3 si 4 de la prima parte?

[tex]5*3^{x+1}=432-3^x=\ \textgreater \ 5*3^x*3=432-3^x \\ Notam:3^x=y,y\ \textgreater \ 0 =\ \textgreater \ 15y=432-y=\ \textgreater \ 16y=432=\ \textgreater \ \\=\ \textgreater \ y=432:16=\ \textgreater \ y=?[/tex]

Egalezi pe 3^x cu y, scrii pe y ca putere de 3 si afli x-ul.

4)Formula combinarilor complementare:
C(2011)(2009)=C(2011)(2011-2009)=C(2011)(2)

C(2012)(2010)=C(2012)(2012-2010)=C(2012)(2)

Aranjamente luate cate 1 este egal cu ce este la indice, adica cu 2011.[tex]C_{2011}^{2}=\frac{2011*2010}{2}\\ C_{2012}^{2}=\frac{2012*2011}{2}[/tex]
Le aduni pe toate trei, pe ultima inmultesti cu 2 ca sa fie la acelasi numitor, apoi le aduni numaratorii si scoti factor comun pe 2011 si faci calculul.

Аноним Аноним · Answer 2 · 2017-04-12T20:05:12+03:00

[tex]\displaystyle \mathtt{3)~5 \cdot 3^{x+1}=432-3^x\Rightarrow 5 \cdot 3^{x+1}+3^x=432\Rightarrow 5 \cdot 3^x \cdot 3+3^x=432\Rightarrow}\\ \\ \mathtt{\Rightarrow 3^x(15+1)=432\Rightarrow3^x \cdot 16=432\Rightarrow3^x= \frac{432}{16} \Rightarrow 3^x=27\Rightarrow }\\ \\ \mathtt{\Rightarrow 3^x=3^3 \Rightarrow x=3}[/tex]

[tex]\displaystyle \mathtt{4)~C_{2011}^{2009}-C_{2012}^{2010}+A_{2011}^1=}\\ \\ \mathtt{ =\frac{2011!}{(2011-2009)!\cdot 2009!}- \frac{2012!}{(2012-2010)!\cdot 2010!}+2011=}\\ \\ \mathtt{= \frac{2011!}{2! \cdot 2009!}- \frac{2012!}{2!\cdot2010!}+2011 =}\\ \\ \mathtt{ =\frac{2009! \cdot 2010 \cdot 2011}{2! \cdot 2009!} - \frac{2010! \cdot 2011 \cdot 2012}{2! \cdot 2010!}+2011=}\\ \\ \mathtt{= \frac{2010\cdot2011}{1\cdot2}- \frac{2011\cdot2012}{1\cdot2}+2011=}[/tex]
[tex]\displaystyle \mathtt{=1005\cdot2011-2011\cdot1006+2011=2011(1005-1006+1)=}\\ \\ \mathtt{=2011(-1+1)=2011 \cdot 0=0}\\ \\ \mathtt{C_{2011}^{2009}-C_{2012}^{2010}+A_{2011}^1=0}[/tex]