1. **Statement of the problem:** We have three exercises involving algebraic expressions and geometry. --- **Exercise 1:** For any nonzero rational number $a$, determine if the following statements are true or false with justification: (a) The product $4a$ is negative. (b) The product of $a$ and its opposite is positive. (c) The product $a \times a$ is negative. (d) Twice the number $a$ is positive. **Step-by-step:** 1. (a) The sign of $4a$ depends on $a$. Since $4$ is positive, $4a$ has the same sign as $a$. So if $a$ is positive, $4a$ is positive; if $a$ is negative, $4a$ is negative. Therefore, the statement "The product $4a$ is negative" is **false** in general. 2. (b) The product of $a$ and its opposite $-a$ is $a \times (-a) = -a^2$. Since $a^2$ is always positive (square of any nonzero number), $-a^2$ is always negative. So the product is negative, not positive. The statement is **false**. 3. (c) The product $a \times a = a^2$ is always positive for any nonzero $a$. So the statement "The product $a \times a$ is negative" is **false**. 4. (d) Twice the number $a$ is $2a$. Its sign depends on $a$. If $a$ is positive, $2a$ is positive; if $a$ is negative, $2a$ is negative. So the statement "Twice the number $a$ is positive" is **false** in general. --- **Exercise 2:** Calculate the following expressions showing all steps: $$A = 1 - \frac{12}{2} + 4 \times (-5)$$ $$B = -3 + 4 - 7 - 6 + 3 \times 4$$ **Step-by-step:** 1. Calculate $A$: $$A = 1 - 6 + 4 \times (-5) = 1 - 6 - 20 = (1 - 6) - 20 = -5 - 20 = -25$$ 2. Calculate $B$: First simplify multiplication: $$3 \times 4 = 12$$ Then substitute: $$B = -3 + 4 - 7 - 6 + 12$$ Calculate stepwise: $$(-3 + 4) = 1$$ $$1 - 7 = -6$$ $$-6 - 6 = -12$$ $$-12 + 12 = 0$$ So, $B = 0$. --- **Exercise 3:** Calculate and simplify the following rational expressions showing all steps: $$E = -\frac{7}{50} + \frac{2}{25} \times \frac{4}{3}$$ $$F = \frac{3}{16} - \frac{1}{6} \div 2$$ $$G = -2 \left( \frac{3}{4} - \frac{2}{-5} \right) + \frac{3}{2}$$ $$H = \frac{7}{-8} + \frac{5}{4} - 1$$ **Step-by-step:** 1. Calculate $E$: $$\frac{2}{25} \times \frac{4}{3} = \frac{8}{75}$$ Find common denominator for $-\frac{7}{50}$ and $\frac{8}{75}$: LCM of 50 and 75 is 150. Convert: $$-\frac{7}{50} = -\frac{21}{150}$$ $$\frac{8}{75} = \frac{16}{150}$$ Sum: $$E = -\frac{21}{150} + \frac{16}{150} = -\frac{5}{150} = -\frac{1}{30}$$ 2. Calculate $F$: Division by 2 is multiplication by $\frac{1}{2}$: $$\frac{1}{6} \div 2 = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$$ Then: $$F = \frac{3}{16} - \frac{1}{12}$$ Find common denominator 48: $$\frac{3}{16} = \frac{9}{48}$$ $$\frac{1}{12} = \frac{4}{48}$$ Subtract: $$F = \frac{9}{48} - \frac{4}{48} = \frac{5}{48}$$ 3. Calculate $G$: Inside parentheses: $$\frac{3}{4} - \frac{2}{-5} = \frac{3}{4} + \frac{2}{5}$$ Find common denominator 20: $$\frac{3}{4} = \frac{15}{20}$$ $$\frac{2}{5} = \frac{8}{20}$$ Sum: $$\frac{15}{20} + \frac{8}{20} = \frac{23}{20}$$ Multiply by $-2$: $$-2 \times \frac{23}{20} = -\frac{46}{20} = -\frac{23}{10}$$ Add $\frac{3}{2} = \frac{15}{10}$: $$G = -\frac{23}{10} + \frac{15}{10} = -\frac{8}{10} = -\frac{4}{5}$$ 4. Calculate $H$: $$\frac{7}{-8} = -\frac{7}{8}$$ Convert all to denominator 8: $$-\frac{7}{8} + \frac{5}{4} - 1 = -\frac{7}{8} + \frac{10}{8} - \frac{8}{8}$$ Sum: $$(-7 + 10 - 8)/8 = (-5)/8 = -\frac{5}{8}$$ --- **Exercise 4 (Geometry):** Given an acute angle $xOy$, points $A$ and $B$ symmetric with respect to $[Ox]$, points $C$ and $D$ on $[Oy]$ such that $OA = OC = OB$ and $OD = OD$. 1. Construct a suitable figure. 2. Prove triangles $OBC$ and $ODA$ are congruent. 3. Let $I$ be the intersection of lines $(AD)$ and $(BC)$. Prove triangles $ICD$ and $IAB$ are congruent. **Step-by-step (proof outlines):** 1. Since $A$ and $B$ are symmetric about $Ox$, $OA = OB$ and $A$ is the reflection of $B$ across $Ox$. 2. Triangles $OBC$ and $ODA$ share $O$ and have $OB = OA$, $OC = OD$ by given, and angles at $O$ equal because of symmetry. By SAS criterion, triangles $OBC$ and $ODA$ are congruent. 3. Point $I$ is intersection of $AD$ and $BC$. Using properties of congruent triangles and symmetry, triangles $ICD$ and $IAB$ are congruent by appropriate criteria (e.g., SAS or ASA) considering corresponding sides and angles. --- **Final answers:** - Exercise 1: All statements (a), (b), (c), (d) are false in general. - Exercise 2: $A = -25$, $B = 0$. - Exercise 3: $E = -\frac{1}{30}$, $F = \frac{5}{48}$, $G = -\frac{4}{5}$, $H = -\frac{5}{8}$. - Exercise 4: Proofs as outlined above.