हल कीजिए:
`(x+1) (x+2) (x+3) (x+4)+1=0, x in R`

(x+1)/3le(2x-1)/4,x in R

Let f (x) = (x ^(3) -4)/((x-1)^(3)) AA x ne 1, g (x)== (x ^(4) -2x ^(2))/(4) AA x in R, h (x) (x ^(3) +4)/((x+1)^(3)) AA x ne -1,

f: R^(+) to R defined by f(x) =2^(x) , x in (0,1), f(x) =3^(x), x in [1, infty) is:

1/(x+1)-1/(x+2)-1/(x+3)+1/(x+4)=0 then x=

Without expanding, find the value of: (i) (x + 1)^4 - 4(x + 1)^3 (x - 1) + 6(x + 1)^2 (x - 1)^2 - 4(x + 1) (x - 1)^3 + (x -1)^4 (ii) (2x - 1)^4 + 4(2x - 1)^3 (3 - 2x) + 6(2x - 1)^2 (3 - 2x)^2 + 4(2x - 1) (3 - 2x)^3 + (3 - 2x)^4

3/(2x)+1/3ge2/(3x)+1/2(x gt0),x in R

Solve : 2|3-x||+3|x-1|=4, x in R

Solve : 2|3-x||+3|x-1|=4, x in R