Given that x satisfies √ (99-x) (x-99) = √ 99-x · √ x-99, y is the integral part of √ 2007 + X, find the value of √ x + y

Given that x satisfies √ (99-x) (x-99) = √ 99-x · √ x-99, y is the integral part of √ 2007 + X, find the value of √ x + y

Under the root sign, 99-x > = 0, x = 0, x > = 99
So x = 99
45²

The result of simplifying 1 + X + X (1 + x) + X (1 + x) ^ 2 +... + X (1 + x) ^ 2008 is

1+x+x(1+x)+x(1+x)^2+...+x(1+x)^2008
=1+[x+x(1+x)+x(1+x)^2+...+x(1+x)^2008]
=1+x[1+(1+x)+...+(1+x)^2008]
=1+x*[(1+x)^2009-1]/(x-1)
=[x*(1+x)^2009-1]/(x-1)

Given 1 + X + x ^ 2 + x ^ 3 = 0, find 1 + X + x ^ 2 + x ^ 3 + +Value of x ^ 2008

1 + X + x ^ 2 + x ^ 3 = 0, 1 + X + x ^ 2 (1 + x) = 0, (1 + x) (1 + x ^ 2) = 0, 1 + x = 0, or 1 + x ^ 2 = 0, x = - 1 1 + X + x ^ 2 + x ^ 3 + --- + x ^ 2008 = 1-1 + 1-1 +... + 1 = 1, students adopt my ha