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ok, sorry,
for the
a^10+a^5+1, you are suppose to factor as the product of two polynomials, such as:
(a^3+2a^2-1)(a^7+1) -- which is not the answer of course
as for the 1^2+2^2+...+n^2, you must first realize, 1^2+2^2+...+n^2=n(n+1)(2n+1)
so now the question becomes:
(n+1)(2n+1)/6=k^2, find the smallest k such that k>1 and n,k are both integers.
i hope this makes easier.
for the whole number question:
(1+1/n)^2=1+2/n+1/n^2
(1+2/(3n))^3=1+2/n+4/(3n^2)+8/(27n^3)
so for every natureal number n,
(1+2/(3n))^3>(1+1/n)^2
thus
1+2/(3n)>(1+1/n)^(2/3)
multiply by n^(2/3) yields:
n^(2/3)+2/(3n(1/3))>(n+1)^(2/3)
and finally:
1/n^(1/3)>3/2((n+1)^(2/3)-n^(2/3))
this is the first half of the solution.
*************Bugpodder,
sorry, u 're wrong again,using derivitve is not good,i'm still waiting for mmiill....
for the second,i just show u the proof of cosx+sinx=sqrt(2)cos(x-pi/4)
=>3^1/2*2^1/2cos(x-pi/4)+2^1/2sin(x-pi/4)=sqrt(2)
<=>sqrt(6)cos(x-pi/4)+sqrt(2)sin(x-pi/4)=sqrt(2)
(acosx+bsinx=c,a^2+b^2>=c^2)
<=>(sqrt(3)/2)cos(x-pi/4)+(1/2)sin(x-pi/4)=(1/2)(divide both side by 2/sqrt(2))
cospi/6cos(x-pi/4)+sinpi/6sin(x-pi/4)=1/2
<=>cos(pi/6-x+pi/4)=cos(pi/3)
<=>cos(x-5pi/12)=cox(pi/3)
=>x-5pi/12=+-pi/3+k2pi
now u can get x (my finger need a rest).
*161 is so easy,160 is too now
*a^10+a^5+1 cause i dont understand what FACTOR is in VNese so i'm wrong but
now its ok...
********mmiill,
i'm rerally sorry for misunderstanding you.please dont be sad,we're still good friennd,ok?
the second problem is too difficult to read,mind if u write it more clearly?
i dont think the third is completely right,u didnt prove that x=1 is the root(we can see
x=1 all along)
wat about the first?
You're really very good at math,have u got any secret?
*************
WE ARE REALLY SO TIRED TO SOLVE SUCH HARD PROBLEM BUT
IF THERE'S A PRIZE ,IT WILL BE MORE INTERESTING,WONT IT?
I SUGGEST THAT :
EACH OF US WILL POST 3 PROBLEM,THE MAN WHO HAVE MOST RIGHT
SOLUTIONS ,WILL GET SOMETHING FROM THE OTHER,OK?
IF U AGREE ,TELL ME WAT THE PRIZE IS!!!
i must go to church now,see u later
:D hey,bugpodder, in 160,is I the incentre of the triangle and the center of inscribe circle?
lol i don't think i can ever mail anything to Viet Nam cuz i am out of $$$. well, let me tell you this, if you can solve these problems, the prize is of course, knowledge and experience, which enables you to do better in math. for me its practice for math contests (sorry Phan not online contests) i write every year. i mean its always good to learn some new techniques. lol i never knew cosx+sinx=sqrt(2)(cos(x-pi/4)) -- thats one hell of a nifty trick! and plus i can use some amusement in math lol! for one i like to see problems that uses trick like that!
ok, factor x^2+3x+2 would be come (x+1)(x+2)
both x+1 and x+2 have integral co-efficients. also there is no trailing term, so if you factor the above expression into (x+1)(x+1)+x+1, that would be wrong because you are adding something in the end.
LOL
prize ?? mmmmm i think it is interesting.
i dont know how it is in VIETNAM ,but i know how things works in CANADA.i have a friend in CANADA and he works like a PICA guy and he is payed about 1000$ for a mounth.
listen now in SERBIA it is catastrophe.a mounth pay in my country is about 100$ for people whit faculty.and that is very small pay. in CANADA it is about 15.000$ for people whit faculty.
and of course internet is so expensive.
im kiding about prize. :D
VIETNAM:
about our conflict it is OK :D . about third problem i know and i seid that it is not good,but i saw that when i finished writeing so i put that solution anyway.about first i havent looked till now but i will.
u said that u have to go to the church.it is very funy for me .what are u ortodox(hmmh i think u are not) or roman or muslim or maybe something else..(do u have a bais eyes like chinamen or ... :D)
CANADA:
thanks for a little help.
SEE YA
:D :D :D :D :D :p :p :p :p :p
a ... .. .. . . b . .. .. . .c ..... . . . . . . . . 3
------ + -------- + ---------- > 3 --------------------------------------
b+c . .. c+a . ... . a+b .. .... . . . (b+c)/a+(a+c)/b+(a+b)/c
this is becouse ( (a+b)/2 > 2/(1/a+1/b) )
aritmetic and ........... {disparity} i dont know on english
a1+a2+a3+......+an . . . .. . . n
-------------------------- > -------------------------
. .. . . . . .n . .. . . . . . . . ..1/a1+1/a2+.....+1/an
where 1,2,3, are indexs
b/a+c/a+a/b+c/b+a/c+b/c>6 becouse
{
(a-b)^2>0
=>
a^2-2ab+b^2>0
=>
a^2+b^2
------------- >2
ab
=>
b/a+a/b >2 (*)
(b/a)+[c/a]+(a/b)+{c/b}+[a/c]+{b/c}>6
becouse (*)
(b/a)+(a/b)+[c/a]+[a/c]+{c/b}+{b/c}> 2+2+2=6
}
now
. 9 . . . . . . . .. .. . ... .... . .. . .. . . . 9 .. .. . . . 3
------------------------------ > ---------------- = ------
(b+c)/a+(a+c)/b+(a+b)/c . . . . .6 .. . .. . . . .2
CANADA:
yes ive been looking for that formula since u gave that problem ,but i think u forgot to DIV whit 6.it is not just n(n+1)(2n+1) check it!
ok, here is the solution (i did forgot to divide by 6):
after you simiplify and rearrange, you get:
(4n+3)2=48n2+1
let x=4n+3
x2-48k2=1
so (x2-48k2)2=x4-96(kx)2+(48k2)2
=(x2+48k2)2-96(2xk)2
so since (7,1) satisfies x^2-48k^2=1
(1,7) (14,97)
then
(97,14) also satisfies the equation
but when x=97, n is not a integer
so similarly using the method above, we find:
(1351,195)
so n=(1351-3)/4=337 -- hehe told you its pretty hard
this is called the AM-GM-HM inequalityQuote:
a1+a2+a3+......+an . . . .. . . n
-------------------------- > -------------------------
. .. . . . . .n . .. . . . . . . . ..1/a1+1/a2+.....+1/an
The Arithmatic mean of a set of numbers is greater or equal to the Geometric Mean which is greater or equal the Harmonic Mean.
mm used AM>=HM
AM of a1,a2,a3,...,an=>(a1+a2+...+an)/n
GM => (a1*a2*...*an)^(1/n)
HM => n/(1/a1+1/a2+...+1/an)
let me write mm's solution more clearly
a/(b+c)+b/(c+a)+c/(a+b)>=3/2
By AM-HM
(a/(b+c)+b/(c+a)+c/(a+b))/3>=3/((b+c)/a+(c+a)/b+(a+b)/c)
a/(b+c)+b/(c+a)+c/(a+b)>=9/(b/a+c/a+c/b+a/b+a/c+b/c)
a/b+b/a=(a2+b2)/ab=(a2+b2-2ab+2ab)/ab=((a+b)2+2ab)/ab>2
so
a/(b+c)+b/(c+a)+c/(a+b)>=9/([b/a+a/b]+[c/a+a/c]+[c/b+b/c])
a/(b+c)+b/(c+a)+c/(a+b)>=9/6
a/(b+c)+b/(c+a)+c/(a+b)>=3/2
Ok ,thats just my idea
The mm'solution for 2 is so not too bad.What about the first?
hey,mmiill,y do u its funny to go to church?i'm roman catholic,
what 'bout u,muslim?
hahahahahahha
no im not muslim im ortodox ,just like russian.i dont like muslim becouse we had a lot of wars against muslim ,in bosnia,in kosovo,and u know for 11. september and for israel .............
im noe saying that there is not good one ,but most of they are fanatic ,and {suicider} and i hate them,becuse my country,s been in crisis becouse of them for 10 years,and still we have prombles in south SERBIA and KOSOVO..
olso i want to say im sorry for people who are muslims and who are good!!!!!
one of my best friend is muslim ,and there is no problem about that :D
about math problems i havent much time and i naveny seen for 2 days
CANADA:
are u shure about 1^2+2^2+........... ? that there is result!!!??? im just asking this :D
not too bed ??????????????????????????
it is same like CANADA guy said!!!!!!!!!!!
SEE YA
i gave u the solution, it is 4 posts up, mmiill.
mmiill,u are communist?cuz u r not catholic,hahahaha...
i hate the bad muslim ,the terrorist.
ignore it!!
hey,bugpodder,y dont u answer for my last post?u dont want me to help u solve 161? do u want the solution for 160?its very easy
lol it slipped my mind. but sure!
IM NOT COMMUNIST!!!!!!!
WHAT ARE U TALKING ABOUT ??????????????????
:mad: :mad: :mad: :mad: :mad:
:D i dont like communist.ok,forget it,lets get backto the maths,cant u solve the first? if not,i'll show u my sol
still npothing ihavent tried yey.but im sorry i cant tolk anymore whit u ,but nice to meet u.
SEE YA
what about whit sqrt 90 (1/10) for the firs problem PHAN.....