The number of real roots of the equation, $$e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$$ is:

We start from the equation

$$e^{4x}+e^{3x}-4e^{2x}+e^{x}+1=0.$$

The exponential function is always positive, so we introduce the substitution

$$t=e^{x}\qquad\text{with}\qquad t>0.$$

Under this change of variable every power of $$e^{x}$$ becomes the corresponding power of $$t$$, giving

$$t^{4}+t^{3}-4t^{2}+t+1=0,\qquad t>0.$$

We now look for roots of this quartic polynomial. A quick test of easy values shows

$$t=1:\;1+1-4+1+1=0,$$

so $$t=1$$ is a root. Therefore the factor $$t-1$$ divides the polynomial. Carrying out the division (or synthetic division) on the coefficients $$1,\,1,\,-4,\,1,\,1$$ we obtain

$$t^{4}+t^{3}-4t^{2}+t+1=(t-1)\bigl(t^{3}+2t^{2}-2t-1\bigr).$$

Again we test the cubic factor at $$t=1$$:

$$1+2-2-1=0,$$

so $$t=1$$ is a root of the cubic as well. Dividing the cubic by $$t-1$$ once more gives

$$t^{3}+2t^{2}-2t-1=(t-1)\bigl(t^{2}+3t+1\bigr).$$

Hence the original quartic completely factors as

$$t^{4}+t^{3}-4t^{2}+t+1=(t-1)^{2}\,(t^{2}+3t+1).$$

We now examine each factor for positive roots.

1. From $$(t-1)^{2}=0$$ we get $$t=1$$, which is clearly positive. Although the factor is squared (so $$t=1$$ is of multiplicity two), it still gives only one distinct positive root.

2. From $$t^{2}+3t+1=0$$ we use the quadratic‐formula statement

$$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a},\quad a=1,\,b=3,\,c=1.$$ Thus

$$t=\frac{-3\pm\sqrt{9-4}}{2}=\frac{-3\pm\sqrt5}{2}.$$

The value $$\dfrac{-3+\sqrt5}{2}\approx -0.382$$ is negative, and $$\dfrac{-3-\sqrt5}{2}$$ is even more negative. Hence the quadratic supplies no positive roots.

Because the substitution $$t=e^{x}$$ restricts us to positive $$t$$, only the value $$t=1$$ survives. Translating back,

$$t=1\;\Longrightarrow\;e^{x}=1\;\Longrightarrow\;x=0.$$ So exactly one real value of $$x$$ satisfies the given equation.

Hence, the correct answer is Option A.