pRoVe that for three numbers a,b,c ,d       (a + b + c + d) >  4 .  (abcd)^1/4?
374695
its so easy ,\n   \n    we know that Arithmetic mean of n no.s > geometric mean of n no.s\n  \n there for four no.s a b c d\n  its AM is (a+b+c+d)/4\n  and its GM is  (abcd)^1/4\n\n  there fore \n    AM > GM\n   ie\n   (a+b+c+d) >  4 . (abcd)^1/4\n  \n        HENCE PROVED ,\n   AND PLEASE DONT ASK SUCH STUPID QUESTIONS\n                !!!!!!  STUPID PEOPLE STUPID QUESTIONS !!!!!\n\nBECAUSE GENIUS LIKE EINSTEIN NAD NEWTO ARE STILL ALIVE IN THIS WORLD
367524
using principle of mathematical induction u can prove this\n check for n=1  2^2 -3 -1=0 which is div by 9\nlet this be true for n=m\n2^2m -3 m -1 =9k  ,where k is any integer\n\nnow for n=m+1 \n2^2(m+1) -3(m+1) -1\n=4.2^2m-3m-4\n=(2^2m-3m-1)+3.2^2m -3\n=9k+3(2^2m-1), now (2^2m -1) is an odd integer divisible by 3 for all m >0\ntherefore the above expression becomes 9k +3.3g ,where g is any integer\n=9(k+g)=9i  where i is an integer \ntherefore the condition for divisibility by 9 is true for n=m+1\nhence by principle of mathematical induction it is true for all n>=1
1051369
Given a+b+c=0\na+b=-c\nCubing both sides, we get\n(a+b)^3=(-c)^3\na^3+3a^2 b+3ab^2+b^3=-c^3\na^3+b^3+c^3=-3a^2 b-3ab^2=-3ab(a+b)=-3ab(-c)=3abc\nTherefore a^3+b^3+c^3=3abc
90506
I'll take it as far as I can.\n\n3 - Add all the numbers and if the result is divisible by 3 then the entire number is 9+1+2+0+3+1+2=18 which is divisible by 3.\n4 - The last 2 numbers are divisible by 4. 12 is divisible by 4 so the entire number is.\n6 - It's divisible by 2 and 3.\n8 - If the last 3 numbers are divisible by 8, then the entire number is. 312/8=39. \n9 - Like for 3, except it has to be divisible by 9. The result, from above, is 18 which is divisible by 9.\n12 - If divisible by 3 and 4.\n18 - If divisible by 2 and 9.\n24 - If divisible by 8 and 3.\n36 - If divisible by 4 and 9.\n72 - If divisible by 8 and 9.
810638
b=a+1\nc=a+2\nd=a+3\ne=a+4\nf=a+5\ng=a+6\nh=a+7\na^2+d^2+f^2+g^2\n=a^2+(a+3)^2+(a+5)^2+(a+6)^2\n=4a^2+28a+70\nb^2+c^2+e^2+h^2\n=(a+1)^2+(a+2)^2+(a+4)^2+(a+7)^2\n=4a^2+28a+70\nhence proved
733649
(4)/(x+7) + (x+3)/(x+7)\n=(4 + (x+3))/(x+7)\n=(4 + x + 3)/(x+7)\n=( x +7)/(x+7)\n=1
258820
4x^2 -3x + 2 = 0 \nD = b^2 - 4ac = 9 - 4*2*4= 9- 32 = - 23<0\nThere are no real solutions, but there are 2 complex solutions.
771168
1= B\n2= A\n3= D\n4= D\n5= D\n6= C
406468
D = {x | x < -3 U x > 4 }
555527
3 is A $4,187 and 4 is D $6.08
745882
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
879176
= x(x+3) + 4(x+3)\n= x^2 + 3x +4x + 12\n= x^2 +7x + 12
970698
(a + b)(c + d + e)\n= a(c + d + e) + b(c + d + e)\n= ac + ad + ae + bc + bd + be\n\nOR\n\n(c + d + e)(a + b)\n= c(a + b) + d(a + b) + e(a + b)\n= etc.
145213
Since A=4,\nThere are four forms for B+C+D=13:\n9+3+1\n8+3+2\n7+5+1\n6+5+2\n\n\nThere are five forms for G+H+I=17:\n9+5+3 [Not valid 5 and 3 are already used for B+C+D]\n9+6+2  \n9+7+1 \n8+7+2 \n8+6+3\n \n\n\nPosibble values for (the order could vary)\nB+C+D and G+H+I then E+F\n9+3+1 and 8+7+2 then 5+6\n8+3+2 and 9+7+1 then 5+6\n7+5+1 and 9+6+2 then 3+8\n7+5+1 and 8+6+3 then 2+9\n6+5+2 and 9+7+1 then 3+8\n\nFirst conclussion E+F=11\nThen D+G = 6\n4+2 [Not Valid, A=4]\n3+3 [Not Valid...]\nValues of D and G  should be 5 and 1\n\nSumarizing:\n\nA = 4\nB+C+D = 6+5+2\nG+H+I = 1+7+9\nE+F = 3+8\n\nTHEN:\n D=5\n H=1
403326
1+2+3+4+5+6+7+8+9/3+8*2-4\n=36+3+16-4\n=51
569945
1433\n1+4 = 5\n3+3 = 6\n5 is one less than 6\n4 is three more than 1
652975
(x+4)/[(3x-1)(3x+1)]-\n(x-2)/(3x+1)=\n\n[(x+4)-(3x-1)(x-2)]/\n[(3x-1)(3x+1)]=\n\n[x+4-3x^2+6x+x-2]/\n[9x^2-1]=\n\n[-3x^2+8x+2]/[9x^2-1]