3(2^(x)+1)-2^(x+2)+5=0

If [1" "x" "1] [(1,2,3),(0,5,1),(0,3,2)][(x),(1),(-2)]=0 then find the value of x.

If [(1,x,1)][(1,3,2),(0,5,1),(0,3,2)][(x),(1),(-2)]=[0] then x is

lim_( x to 0) ((1+5x^(2))/(1+3x^(2)))^(1//x^(2)) = … .

lim_(x rarr0)((1+5x^(2))/(1+3x^(2)))^(1/x^(2))=

x^2-2x+2=0"( 2)"x^2-3x+3=0x^2-3x+4=0"( 4)"x^2-5x+5=0 A quadratic equation whose roots are cosec20 andsec 0 can be(1) x2-2x+ 2 =0(2) x2-3x + 3 = 0(3) x2-3x+ 4= 0(4) x-5x+5= 0

lim_(x rarr0)((5x^(2)+1)/(3x^(2)+1))^(1/x^(2))

Let the equation x^(5) + x^(3) + x^(2) + 2 = 0 has roots x_(1), x_(2), x_(3), x_(4) and x_(5), then find the value of (x_(2)^(2) - 1)(x_(3)^(2) - 1)(x_(4)^(2) - 1)(x_(5)^(2) - 1).

Solve the following equations where x in R a) (x-1)|x^(2)-4x+3|+2x^(2)+3x-5=0 b) |x^(2)+4x+3|+2x+=0 c) |x+3|(x+1)+|2x+5|=0