1. **Problem statement:** Given the inequalities $4 < y - x < 9$ and $10 < x^2 + y^2 < 45$, and the functions $B = \frac{x+1}{2}$ and $A = \sqrt{\frac{x^2 + y^2}{y - x}}$, we need to find the bounds (حصر) for $A$ and $B$. 2. **Step 1: Analyze the bounds for $A$** - From the inequalities, $y - x$ is between 4 and 9, so $4 < y - x < 9$. - Also, $x^2 + y^2$ is between 10 and 45, so $10 < x^2 + y^2 < 45$. - Since $A = \sqrt{\frac{x^2 + y^2}{y - x}}$, substitute the bounds: $$A_{min} = \sqrt{\frac{10}{9}} = \sqrt{\frac{10}{9}} = \frac{\sqrt{10}}{3} \approx 1.054$$ $$A_{max} = \sqrt{\frac{45}{4}} = \sqrt{11.25} = 3.354$$ - Therefore, the bounds for $A$ are: $$1.054 < A < 3.354$$ 3. **Step 2: Analyze the bounds for $B$** - $B = \frac{x+1}{2}$. - To find bounds for $B$, we need bounds for $x$. - From $10 < x^2 + y^2 < 45$ and $4 < y - x < 9$, the exact bounds for $x$ are not directly given, but since $x^2 + y^2 < 45$, $|x| < \sqrt{45} = 6.708$. - Also, $y - x > 4$ implies $y > x + 4$. - To find the minimum and maximum of $B$, consider the range of $x$ within the constraints. - Since $x$ can vary approximately between $-6.7$ and $6.7$, then: $$B_{min} = \frac{-6.7 + 1}{2} = \frac{-5.7}{2} = -2.85$$ $$B_{max} = \frac{6.7 + 1}{2} = \frac{7.7}{2} = 3.85$$ - So the bounds for $B$ are approximately: $$-2.85 < B < 3.85$$ 4. **Step 3: Analyze $BJ = x + \frac{1}{2}$ and compare $BJ16$, $B$, and $B-8$** - $BJ = x + 0.5$. - Since $B = \frac{x+1}{2}$, then $2B = x + 1$. - Comparing $BJ16$, $B$, and $B-8$ requires more context, but generally: - $BJ16$ likely means $BJ$ evaluated at 16 or some notation; assuming $BJ16 = 16 + 0.5 = 16.5$. - $B$ varies as above. - $B - 8$ shifts $B$ down by 8. - Without more context, the comparison is: $$BJ16 = 16.5 > B_{max} = 3.85 > B_{min} - 8 = -2.85 - 8 = -10.85$$ 5. **Step 4: Evaluate $A = |6 - 27| - |\sqrt{5} - 2| + \sqrt{(\sqrt{5} - 4)^2}$** - Calculate each term: $$|6 - 27| = | -21 | = 21$$ $$|\sqrt{5} - 2| \approx |2.236 - 2| = 0.236$$ $$\sqrt{(\sqrt{5} - 4)^2} = |\sqrt{5} - 4| = |2.236 - 4| = 1.764$$ - Sum: $$A = 21 - 0.236 + 1.764 = 22.528$$ 6. **Step 5: Absolute value expression analysis** - Expression: $$|\alpha^2 - \alpha| + \sqrt{(1 - \alpha)^2} - |1 - \alpha^2| + 1$$ with $10 < \alpha < 1$ is contradictory since $10 < \alpha < 1$ is impossible; likely a typo. 7. **Step 6: Analyze $A(x) = |-3x + 4| + |-5 + x| - 2$ for real $x$** - (1) Write $A(x)$ without absolute value symbols by considering cases: - Find points where expressions inside absolute values change sign: - $-3x + 4 = 0 \Rightarrow x = \frac{4}{3} \approx 1.333$ - $-5 + x = 0 \Rightarrow x = 5$ - Cases: - For $x < \frac{4}{3}$: $$|-3x + 4| = 4 - 3x$$ - For $x \geq \frac{4}{3}$: $$|-3x + 4| = 3x - 4$$ - For $x < 5$: $$|-5 + x| = 5 - x$$ - For $x \geq 5$: $$|-5 + x| = x - 5$$ - Combine cases: - For $x < \frac{4}{3}$ and $x < 5$ (i.e., $x < \frac{4}{3}$): $$A(x) = (4 - 3x) + (5 - x) - 2 = 7 - 4x$$ - For $\frac{4}{3} \leq x < 5$: $$A(x) = (3x - 4) + (5 - x) - 2 = 2x - 1$$ - For $x \geq 5$: $$A(x) = (3x - 4) + (x - 5) - 2 = 4x - 11$$ - (2) Expression without absolute values is piecewise linear as above. - (3) Solve $A(x) = 12$: - Case 1: $7 - 4x = 12 \Rightarrow -4x = 5 \Rightarrow x = -\frac{5}{4} = -1.25$ (valid since $x < \frac{4}{3}$) - Case 2: $2x - 1 = 12 \Rightarrow 2x = 13 \Rightarrow x = 6.5$ (not valid since $x < 5$ in this case) - Case 3: $4x - 11 = 12 \Rightarrow 4x = 23 \Rightarrow x = 5.75$ (valid since $x \geq 5$) - Solutions are: $$x = -1.25 \quad \text{and} \quad x = 5.75$$ **Final answers:** - $1.054 < A < 3.354$ - $-2.85 < B < 3.85$ - $A = 22.528$ (evaluated expression) - $A(x) = 12$ solutions: $x = -1.25$ and $x = 5.75$