x3 + x2 - x - 1 = x2(x+1) - 1(x+1) =(x+1)(x2-1) = (x+1)(x+1)(x-1)
(x - 1)/x = x/x - 1/x = 1 - 1/x
(x - 1)(x - 1)(x - 1) - (x + 1)(x + 1)(x + 1)
Note that x² - 1 can be written as: x² + x - x - 1 Then, factor-by-group the expression to get: x(x + 1) - (x + 1) = (x - 1)(x + 1) Hence, x2 - 1 = (x + 1)(x - 1)
0
1 to the power of 40 can be expanded to be: 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 X 1 Which would equal ..... 1
1 - x/(x2 + x) = 1 - x/[x(x+1)] = 1 - 1/(x+1) = (x+1)/(x+1) - 1/(x+1) = x/(x + 1)
120 = 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 = 1 The number one to the power of any non-negative integer is 1.
X+1=X=X+1
1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 = 19
-1 x -1 x -1 -x -x -1 -2x - 1
cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x
x3 + x2 - x - 1 = x2(x+1) - 1(x+1) =(x+1)(x2-1) = (x+1)(x+1)(x-1)
(x - 1)/x = x/x - 1/x = 1 - 1/x
x3 + 1 = x3 + x2 - x2 - x + x + 1 = x2(x + 1) - x(x + 1) +1(x + 1) = (x + 1)(x2 - x + 1)
Well, isn't that a happy little question! To find the derivative of (x-1)^x, we'll need to use logarithmic differentiation. Start by taking the natural logarithm of both sides, then apply implicit differentiation to find the derivative. Remember, there are no mistakes, just happy little accidents in math!
(x - 1)(x - 1)(x - 1) - (x + 1)(x + 1)(x + 1)