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Okay, Victor. Let's tackle these questions in English.

Here's the solution for Question 1:

Find the y-coordinate of the point on the curve where $x=1$ .
Substitute $x=1$ into the equation of the curve $y = x^2 - 3x + 2$ :
$y = (1)^2 - 3(1) + 2$ $y = 1 - 3 + 2$ $y = 0$
So, the point of tangency is $(1, 0)$ .
Find the derivative of the curve to get the slope function.
Differentiate $y = x^2 - 3x + 2$ with respect to $x$ :
$\frac{dy}{dx} = \frac{d}{dx}(x^2) - \frac{d}{dx}(3x) + \frac{d}{dx}(2)$ $\frac{dy}{dx} = 2x - 3$
Evaluate the derivative at $x=1$ to find the slope of the tangent line.
Substitute $x=1$ into the derivative:
$m = 2(1) - 3$ $m = 2 - 3$ $m = -1$
Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$ , to find the equation of the tangent.
Using the point $(x_1, y_1) = (1, 0)$ and the slope $m = -1$ :
$y - 0 = -1(x - 1)$ $y = -x + 1$
The equation of the tangent to the curve at the point where $x=1$ is $\boxed{y = -x + 1}$ .

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