1,565 = x5 1,5651/5 = (x5)1/5 1,5651/5 = x5/5 1,5651/5 = x 4.3541 = x
5+5x5=30 but (5+5)x5=50
1
0 because X= -1
7=x+1 6=x
1,565 = x5 1,5651/5 = (x5)1/5 1,5651/5 = x5/5 1,5651/5 = x 4.3541 = x
5+5x5=30 but (5+5)x5=50
x10 = x5.x5 = (x5)2 [x power five whole squared] Equation is x10 + x5 - 2 Replacing x10 (x5)2+x5 - 2 Substituting x5 by Y Equation becomes Y2 + Y - 2 = 0 Y2 + 2Y - Y - 2 = 0 Y(Y + 2) - 1(Y+2) = 0 (Y-1)(Y+2) = 0 The values of Y are 1 and -2 Y = x5 = 1 Therefore, x = 1 Y = x5 = -2 x = fifth root of -2, which is an imaginary value.
1
0 because X= -1
7=x+1 6=x
Since x and y are variables, the sum of 5x and 5y could be literally any number.
The answer to the problem of 12X plus 13 equals 9 is x equals -1/3.
x = 0
1+x=dd+x so dd=1 so d=1 1+x=89b+x so 89b=1 so b=1/89 bc+x=1+x so bc=1 so c=1/(1/89) = 89 so cd=89 (and x=88)
(1, 9)
5