if `x < y < 2x` and the mean and median of `x,y,2x` are `15,12` respectively then `x`

Using the properties of determinants, solve the following for x |{:(a+x,a-x,a-x),(a-x,a+x,a-x),(a-x,a-x,a+x):}|=0

If 5{x}=x+[x] and [x]-{x}=(1)/(2) when {x} and [x] are fractional and integral part of x then x is

simplify (x+x^(2)+x^(3)+x^(4)+x^(5)+x^(6)+x^(7))/(x^(-3)+x^(-4)+x^(-5)+x^(-6)+x^(-7)+x^(-8)+x^(-9))

The factor of x^8 + x^4 +1 are (A) (x^4 + 1 - x^2), (x^2 +1 +x), (x^2 + 1 - x ) (B) x^4 + 1 -x^2 , (x^2 - 1 + x), (x^2 +1 + x) (C ) (x^4 - 1 + x^2, (x^2 - 1 + x), (x^2 + 1 + x) (D) (x^4 -1 + x^2), (x^2 + 1 - x), (x^2 + 1 +x)

If f(x)=|{:(1,x,x+1),(2x,x(x-1),(x+1)x),(3x(x-1),x(x-1)(x-2),(x+1)x(x-1)):}| then

Using properties of determinants, solve for x:|[a+x, a-x, a-x],[ a-x, a+x, a-x],[ a-x, a-x, a+x]|=0

Using properties of determinants, solve for x:|[a+x, a-x, a-x],[ a-x, a+x, a-x],[ a-x, a-x, a+x]|=0

Using properties of determinants, solve for x:|[a+x, a-x, a-x], [a-x, a+x, a-x], [a-x, a-x ,a+x]|=0