2X + 3 / 4-4-x 6 / 2 + 1 / 2 x

2X + 3 / 4-4-x 6 / 2 + 1 / 2 x


Original formula = 3 / 2 (x + 2) + 6 / (x + 2) (X-2) + 1 / (X-2)
=(3x-6+12+2x+4)/2(x+2)(x-2)
=(5x+10)/2(x+2)(x-2)
=5(x+2)/2(x+2)(x-2)
=5/(2x-4)



How to solve the square of X + 2x + 1 = 6


How to solve the square of X + 2x + 1 = 6
(x+1)^2=6
x+1=±√6
x1=-1+√6
x2=-1-√6



The equation kx2 - (k-1) x + 1 = 0 of X has rational roots, so we can find the value of integer K


(1) When k = 0, x = - 1, the equation has rational roots. (2) when k ≠ 0, because the equation has rational roots, if K is an integer, then △ = (k-1) 2-4k = k2-6k + 1 must be a complete square, that is, there is a non negative integer m, such that k2-6k + 1 = m2