3^2+4^2=5^2,   3^3+4^3+5^3=6^3,4*2^4+7^4+8^4=9^4.   For which k   X1^5+X2^5+.....+Xk^5=Y^5 has integer answes?
962658
3^2+4^2=5^2\n25=25\n\n3^3+4^3+5^3=6^3\n27+64+125 = 216\n\n4*2^4+7^4+8^4=?\n4*16+2401+4096=?\n64+6497=6561\n\nI don´t get the last part.
746280
c^2 = 3^2 + 4^2\nc^2 = 9 + 16 = 25\nc = square root (25) = 5
403326
1+2+3+4+5+6+7+8+9/3+8*2-4\n=36+3+16-4\n=51
636732
8x^2y^3+5y^2 when "FACTORED COMPLETELY"=\n\ny^2 (8yx^2+5). \n\nThis should be right. My dad is a complete genius and he did this in like 2 seconds. I double checked it on my calculator too by plugging in numbers, so I can assure you he's right.\n\nHope this helps! Have a nice day :)
871981
3 to the 5th power\nor 3*3*3*3*3
1157065
(x+5)(2x^2+4x+2)\nx(2x^2+4x+2)+5(2x^2+4x+2)\n2x^3+4x^2+2x+10x^2+20x+10\n2x^3+14x^2+22x+10
798980
Well to solve this we need the identity:\n\nx^(a+b) = x^a*x^b\n\n\n2^(2*k+5)-248 = 2^(2*K)\n\n=> 2^(2*k)*2^5 - 248 = 2^(2*k)\n=> 2^(2*k)*(2^5-1) - 248 = 0\n=> 31*2^(2*k) = 248\n=> 2^(2*k) = 8\n now we know that 2^3 = 8 (to solve this type of equation more generally we would nee dlogarithms, but no need here)\n\nand so finally\n2^(2*k) = 2^3\n=> 2*k = 3\n=> k = 1.5
67692
(4X^3 +12X^2) + (12X +36)= 4X^2(X+3) + 12(X+3)\nX+3 is common.. hence we could write...\nas (X+3)(4X^2 + 12)= 4(x^2+3)(x+3)
1103883
Y=X^5-X^-3+X^2\nY'=5X^4+3X^-4+2X\nY''=20X^3-12X^-5+2
114816
-3+5=5-3y+4y       \n   2=5+y\n  2-5=y \n   -3=y\n because...          2=5-3(-3)+4(-3)\n                     2=5+9-12 \n                     2= 14-12         \n                     2=2
447887
You probably mean\np(x) = 2x^4 +5x^3 – 3x^2 -2x -6\nThe answer is e) Both b,d\nbecause when you multiply the factors you should get the original expression and (x+3)(2x^3 –x^2 -2)= 2x^4 +5x^3 – 3x^2 -2x -6\nThis was quickly seen because you need x^4 which is only obtainable by x multiplied by x^3 in your problem :)\nHope this helps
1043913
Assume we found an integer N where N = x^2 - y^2 where a and b are integers.\n\ncan N^k be expressed in such a form too? Is that what u r asking?\n\nOnly N that can be factored as N=c.d where c and d are both odd, or even, can be expressed as N = x^2 - y^2 for integer x and y.\n\nSo, if N=c.d where c and d r both odds(or even), N^k=c^k.d^k.... and since odd^k is still odd and even^k is still even, N^2 can be expressed as x^2 - y^2 for some integers x and y.
735971
6k^2=k+2\n=>6k^2-k-2=0\n=>6k^2-4k+3k-2=0\n=>2k(3k-2)+1(3k-2)=0\n=>(3k-2)(2k+1)=0\nTherefore,either 3k-2=0 or 2k+1=0\nIf 3k-2=0,then k=2/3 and if 2k+1=0 then k= - 1/2\nkTherfore,k=2/3 or - 1/2 ans
700473
(x^5 + y^5) =  (x+y)(x^4 - x^3*y + x^2*y^2 - x*y^3 + y^4).
817461
a^4 - b^4\n\n= (a^2 + b^2)(a^2 - b^2)\n\n= (a^2 + b^2)(a + b)(a - b)
60270
5+3(18/6) = 5+3(3) = 5+9 = 14\n\n[2(4)+17]*9 = (8+17)*9 = 25*9 = 225
445755
are you simplifying? if so, here's your answer:\n\n(x^2-5x+6)(x^2-5x+5)=\nx^4-5x^3+5x^2-5x^3+25x^2-25x+6x^2-30x+30=\nx^4-10x^3+36x^2-55x+30\n\nwtf cranger, i got it right! what do you mean by CORRECT? check the bottom line