1. **Problem 1:** Find the product of $(x + 5)$ and $(9 - y + z)$. Formula: Use distributive property: $(A + B)(C + D + E) = A(C + D + E) + B(C + D + E)$. Work: $$(x + 5)(9 - y + z) = x(9 - y + z) + 5(9 - y + z)$$ $$= 9x - xy + xz + 45 - 5y + 5z$$ 2. **Problem 2:** Expand $(a + b)(a + b)$. Formula: $(a + b)^2 = a^2 + 2ab + b^2$. Work: $$(a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2$$ 3. **Problem 3:** Which is larger, $25 imes 78$ or $26 imes 77$? Calculate: $$25 imes 78 = 1950$$ $$26 imes 77 = 2002$$ Answer: $26 imes 77$ is larger. 4. **Problem 4:** Verify if $(4m + 1)^2 - (4m + 5)^2$ is always a multiple of 2. Use difference of squares identity: $$a^2 - b^2 = (a - b)(a + b)$$ Let $a = 4m + 1$, $b = 4m + 5$: $$(4m + 1)^2 - (4m + 5)^2 = ((4m + 1) - (4m + 5))((4m + 1) + (4m + 5))$$ $$= (-4)(8m + 6) = -32m - 24$$ Since $-32m - 24$ is divisible by 2 for all integers $m$, the expression is always a multiple of 2. 5. **Problem 5:** Using identity find $145 imes 135$. Use identity: $$(x + a)(x - a) = x^2 - a^2$$ Rewrite: $$145 imes 135 = (140 + 5)(140 - 5) = 140^2 - 5^2 = 19600 - 25 = 19575$$ 6. **Problem 6:** Choose three consecutive numbers, square the middle one and subtract the product of the other two. Repeat with other sets. What pattern? Let the three consecutive numbers be $n-1$, $n$, $n+1$. Expression: $$n^2 - (n-1)(n+1)$$ Expand: $$n^2 - (n^2 - 1) = n^2 - n^2 + 1 = 1$$ Pattern: The result is always 1. Algebraic equation: $$n^2 - (n-1)(n+1) = 1$$ Expanded and verified as true identity.