a(x5) + b(x5) + c(x5) + d(x5) + e(x5) = abcde(a+b+c+d+e) x5 = abcdeThis equation has at least 5 variables. To solve for all of them requires at least 4 more equations.
x5 is an expression, not an equation nor an inequality. An expression cannot be solved.
is a quintic expression in x (NOT an equation).
For the equation: x5+7x3-30x=0 The highest exponent in the entire equation is 5 (from x5), so the equation is of degree 5.
y-x5 is an expression. An expression is not equivalent to any equation.
7
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.
Assuming that the question is find x in x5=1610; the answer is 4.378901
The additive inverse is x5 + 2x - 2.
5x + 5y = 5(x+y)
x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4
x*-4=8+(2-5*x) =-4x=8+2-x5 x*-4=8+(2-5*x) =-4x=8+2-x5 x*-4=8+(2-5*x) =-4x=8+2-x5