Given functions f(x) = 2x2 − (3x)/(4) and g(x) = (x+1)/(1-2x), find the values of x for which g(x−1) = f(1) and f(x+1) = 0.

Substitute $x=1$ into $f(x)$:

$$f(1) = 2(1)^2 - \frac{3(1)}{4} = 2 - \frac{3}{4} = \frac{8}{4} - \frac{3}{4} = \frac{5}{4}$$

Substitute $x-1$ into $g(x)$:

$$g(x-1) = \frac{(x-1)+1}{1-2(x-1)} = \frac{x}{1-2x+2} = \frac{x}{3-2x}$$

$$\frac{x}{3-2x} = \frac{5}{4}$$

Cross-multiply:

$$4x = 5(3-2x)$$ $$4x = 15 - 10x$$

Add $10x$ to both sides:

$$4x + 10x = 15$$ $$14x = 15$$

Divide by 14:

$$x = \frac{15}{14}$$

The value of $x$ for $g(x-1) = f(1)$ is $\boxed{\frac{15}{14}}$.

Substitute $x+1$ into $f(x)$:

$$f(x+1) = 2(x+1)^2 - \frac{3(x+1)}{4}$$

$$2(x+1)^2 - \frac{3(x+1)}{4} = 0$$

Multiply the entire equation by 4 to eliminate the fraction:

$$8(x+1)^2 - 3(x+1) = 0$$

Factor out the common term $(x+1)$:

$$(x+1)[8(x+1) - 3] = 0$$

Simplify the expression inside the square brackets:

$$(x+1)[8x + 8 - 3] = 0$$ $$(x+1)[8x + 5] = 0$$

This equation holds true if either factor is zero:

$$x+1 = 0 \quad \text{or} \quad 8x+5 = 0$$

Solving the first equation:

$$x = -1$$

Solving the second equation:

$$8x = -5$$ $$x = -\frac{5}{8}$$

The values of $x$ for which $f(x+1) = 0$ are $\boxed{-1, -\frac{5}{8}}$.