Let's break down the problem step by step.
### Step 1: Simplify the Expression [tex]\( y \)[/tex]
We start with the given expression:
[tex]\[ y = x^2 \cdot \left\lvert\, 1 \cdot \frac{x}{x^2} \right\rvert \][/tex]
First, simplify the part inside the absolute value:
[tex]\[ \frac{x}{x^2} = \frac{1}{x} \][/tex]
So, the expression becomes:
[tex]\[ y = x^2 \cdot \left\lvert \frac{1}{x} \right\rvert \][/tex]
The absolute value of [tex]\(\frac{1}{x}\)[/tex] is [tex]\(\frac{1}{|x|}\)[/tex]. Thus:
[tex]\[ y = x^2 \cdot \frac{1}{|x|} \][/tex]
This simplifies further to:
[tex]\[ y = \frac{x^2}{|x|} \][/tex]
Now, we recall that [tex]\(|x|\)[/tex] is always positive. Therefore:
- If [tex]\( x > 0 \)[/tex], then [tex]\(|x| = x\)[/tex] and [tex]\[ \frac{x^2}{x} = x \][/tex]
- If [tex]\( x < 0 \)[/tex], then [tex]\(|x| = -x\)[/tex] and [tex]\[ \frac{x^2}{|x|} = \frac{x^2}{-x} = -x \][/tex]
So, [tex]\( y = x \)[/tex] for [tex]\( x > 0 \)[/tex] and [tex]\( y = -x \)[/tex] for [tex]\( x < 0 \)[/tex].
### Step 2: Evaluate [tex]\( x \cdot H \cdot \frac{1}{x} \cdot y \)[/tex]
Given the expression to find:
[tex]\[ x \cdot H \cdot \frac{1}{x} \cdot y \][/tex]
Let's choose [tex]\( x = 1 \)[/tex] for simplicity:
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 1 \][/tex]
Assuming [tex]\( H \)[/tex] is a constant, say [tex]\( H = 1 \)[/tex]:
[tex]\[ x \cdot H \cdot \frac{1}{x} \cdot y = 1 \cdot 1 \cdot \frac{1}{1} \cdot 1 = 1 \][/tex]
Combining all parts, we derive the overall result:
[tex]\[ \boxed{1} \][/tex]
