Question

Ex 32 si 33!! Va rog multtt

Answer 1

32.
    ___  ___  _____
a+a,b+a,bc+a,bcd =18,503
a+a+0,1b+a+0,1b+0,01c+a+0,1b+0,01c+0,001d=18,503
4a+0,3b+0,02c+0,001d=18,503
d=3
4a+0,3b+0,02c+0,001×3=18,503
4a+0,3b+0,02c=18,503-0,003
4a+0,3b+0,02c=18,50
c=5
4a+0,3b+0,02×5=18,50
4a+0,3b+0,1=18,50-0,1
4a+0,3b=18,4
b=8       
4a+0,3×8=18,4
4a+2,4=18,4
4a=18,4-2,4
4a=16
a=4

33.
a=1/2+1/3+1/4+...1/2012
b=1/2+2/3+3/4+...2011/2012
----------------------------------------
a+b=1/2+1/3+1/4+....+1/2012+1/2+2/3+3/4+2011/2012=
=2/2+3/3+4/4+....+2012/2012=
=1+1+1+.....+1=2011
nr de termeni  2012-2+1=2011    

Answer 2

   
[tex]\displaystyle\\ 32)\\ ~~~~~\boxed{a}~~~~~~~+\\ ~~~~~\boxed{a},\boxed{b}\\ ~~~~~\boxed{a},\boxed{b}\boxed{c}\\ ~~~~~\boxed{a},\boxed{b}\boxed{c}\boxed{d}\\ ---------\\ \boxed{1}\boxed{8},\boxed{5}\boxed{0}\boxed{3}\\\\\\ ~~~~~\boxed{4}~~~~~~~+\\ ~~~~~\boxed{4},\boxed{8}\\ ~~~~~\boxed{4},\boxed{8}\boxed{5}\\ ~~~~~\boxed{4},\boxed{8}\boxed{5}\boxed{3}\\ ---------\\ \boxed{1}\boxed{8},\boxed{5}\boxed{0}\boxed{3}[/tex]


[tex]\displaystyle\\ 33)\\ a= \frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2012}\\\\ b= \frac{1}{2}+ \frac{2}{3}+\frac{3}{4}+\cdots+\frac{2011}{2012}\\\\ _\texttt{\bf Daca analizam numitorii, observam ca fiecare din cele 2 siruri}\\ _\texttt{\bf are acelasi nr. de termeni.}\\\\ \texttt{Calculam numarul de termeni:}\\ n = 2012-2+1 = \boxed{\bf 2011 ~termeni}\\\\ _\texttt{Adunam pe a cu b, adunand termenul 1 de la a cu termenul 1 de la b, }\\ _\texttt{termenul 2 de la a cu termenul 2 de la b, ... samd.}[/tex]


[tex]\displaystyle\\ a+b= \left(\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2012}\right) + \left(\frac{1}{2}+ \frac{2}{3}+\frac{3}{4}+\cdots+\frac{2011}{2012}\right)=\\\\ =\frac{1}{2}+\frac{1}{2}+ \frac{1}{3}+\frac{2}{3}+\frac{1}{4}+\frac{3}{4}+\cdots+\frac{1}{2012}+\frac{2011}{2012}=\\\\ =\frac{2}{2}+\frac{3}{3} +\frac{4}{4}+\cdots+\frac{2012}{2012}= \underbrace{1+1+1+\cdots+1}_{\texttt{2011 de unuri}} = 2011\times1=\boxed{\bf 2011}[/tex]