1. The problem is to identify which of the given functions correspond to the first or second notable identities: $$\alpha^2 + 2\alpha b + b^2 = (a + b)^2$$ $$\alpha^2 - 2\alpha b + b^2 = (a - b)^2$$ 2. Recall the first identity: $a^2 + 2ab + b^2 = (a + b)^2$ and the second identity: $a^2 - 2ab + b^2 = (a - b)^2$. 3. We will check each function to see if it matches either identity by rewriting them in the form $a^2 \pm 2ab + b^2$. 4. For $A(x) = x^2 - 2x + 1$, rewrite as $x^2 - 2 \cdot x \cdot 1 + 1^2 = (x - 1)^2$ which matches the second identity. 5. For $B(x) = 4 + 4x + x^2$, rewrite as $x^2 + 4x + 4 = (x + 2)^2$ which matches the first identity. 6. For $C(x) = 9x^2 - 6x + 1$, rewrite as $(3x)^2 - 2 \cdot 3x \cdot 1 + 1^2 = (3x - 1)^2$ which matches the second identity. 7. For $D(x) = 9 + 12x + 4x^2$, rewrite as $4x^2 + 12x + 9 = (2x + 3)^2$ which matches the first identity. 8. For $E(x) = 81x^2 - 18x + 1$, rewrite as $(9x)^2 - 2 \cdot 9x \cdot 1 + 1^2 = (9x - 1)^2$ which matches the second identity. 9. For $F(x) = 25x^2 - 20x + 4$, rewrite as $(5x)^2 - 2 \cdot 5x \cdot 2 + 2^2 = (5x - 2)^2$ which matches the second identity. 10. For $G(x) = x^2 + x + 1$, it does not fit either identity because the middle term is not twice the product of the square roots of the first and last terms. 11. For $H(x) = 1 - \frac{2}{x} + \frac{1}{9x^2}$, rewrite as $1^2 - 2 \cdot 1 \cdot \frac{1}{3x} + \left(\frac{1}{3x}\right)^2 = \left(1 - \frac{1}{3x}\right)^2$ which matches the second identity. 12. For $I(x) = 9x^2 - 30x + 25$, rewrite as $(3x)^2 - 2 \cdot 3x \cdot 5 + 5^2 = (3x - 5)^2$ which matches the second identity. 13. For $J(x) = 16 - 24x + 9x^2$, rewrite as $9x^2 - 24x + 16 = (3x - 4)^2$ which matches the second identity. Final answers: - First identity: $B(x), D(x)$ - Second identity: $A(x), C(x), E(x), F(x), H(x), I(x), J(x)$ - Neither: $G(x)$