1. Given expressions with absolute values, rewrite without absolute values assuming $\pi \approx 3.14$: a) $|3 - \pi| + |4 - \pi|$ We know $3 - \pi \approx 3 - 3.14 = -0.14 < 0$, so $|3 - \pi| = -(3 - \pi) = \pi - 3$. Similarly, $4 - \pi \approx 0.86 > 0$, so $|4 - \pi| = 4 - \pi$. Sum: $ (\pi - 3) + (4 - \pi) = 1$ b) $|2\pi - 6|$ $2\pi \approx 6.28$, so $2\pi - 6 = 0.28 > 0$, thus $|2\pi - 6|=2\pi - 6$ c) $|1 - \sqrt{2}|$ $\sqrt{2} \approx 1.414$, $1 - 1.414 = -0.414 < 0$, so $|1 - \sqrt{2}|=\sqrt{2} - 1$ d) $|\sqrt{2} - \sqrt{3}| - |\sqrt{3} - \sqrt{2}|$ Since $\sqrt{3} > \sqrt{2}$, $\sqrt{2} - \sqrt{3} < 0$ so $|\sqrt{2} - \sqrt{3}|=\sqrt{3} - \sqrt{2}$ Also $|\sqrt{3} - \sqrt{2}|=\sqrt{3} - \sqrt{2}$ Therefore difference is $0$ 2. For $1 < x < 2$: a) $|x - 1| = x - 1$ since $x - 1 > 0$; $|x - 2| = 2 - x$ since $x - 2 < 0$ b) Evaluate $$B= \frac{|x - 1|}{x - 1} + \frac{|x - 2|}{x - 2} = \frac{x-1}{x-1} + \frac{2-x}{x-2}$$ Since $x-1 > 0$, $\frac{x-1}{x-1} = 1$ Also, $x-2 < 0$, denominator negative, numerator $2 - x = -(x - 2)$ So $\frac{2 - x}{x - 2} = \frac{-(x-2)}{x-2} = -1$ Hence $B = 1 + (-1) = 0$ 3. For $A = |x + 3| - |1 - x|$: a) If $-3 < x < 1$: Then $x+3 > 0$, so $|x+3| = x+3$. Also, $1 - x > 0$ so $|1 - x| = 1 - x$ Thus, $A = (x+3) - (1 - x) = x + 3 - 1 + x = 2x + 2$ b) For $x < -3$: $x+3 < 0$, so $|x + 3|=-(x+3)=-x - 3$ Also, $1 - x > 0$ because $x < -3 < 1$, so $|1 - x|=1 - x$ Then $A = (-x - 3) - (1 - x) = -x - 3 - 1 + x = -4$ So $A$ is constant $-4$ for $x < -3$. c) For $x > 1$: $x + 3 > 0 \Rightarrow |x + 3|=x + 3$ $1 - x < 0 \Rightarrow |1 - x|=-(1 - x) = x - 1$ So $A = (x + 3) - (x - 1) = 4$ Hence $A$ is also constant $4$ for $x > 1$. 4. With $\alpha < \beta < \gamma < 1$: a) $A = |\alpha - \beta| + |\beta - \gamma| + |\gamma - \alpha|$ Since $\alpha < \beta$, $|\alpha - \beta| = \beta - \alpha$ Similarly $|\beta - \gamma| = \gamma - \beta$, $|\gamma - \alpha| = \gamma - \alpha$ Sum: $(\beta - \alpha) + (\gamma - \beta) + (\gamma - \alpha) = (\beta - \alpha + \gamma - \beta + \gamma - \alpha) = 2(\gamma - \alpha)$ b) $B = 2|\beta - \alpha| + |\alpha + \beta - 2\gamma| - 2|\alpha - \beta|$ Note $|\beta - \alpha| = \beta - \alpha$ $B = 2(\beta - \alpha) + |\alpha + \beta - 2\gamma| - 2(\beta - \alpha) = |\alpha + \beta - 2\gamma|$ c) $\Gamma = |\alpha - \beta + \gamma| + | - \alpha + \beta - \gamma| + |\gamma - 1|$ Rewrite $| - \alpha + \beta - \gamma| = |\beta - \alpha - \gamma|$ Exact simplifications depend on numeric values but expressions are ready for substitution. d) $\Delta = \frac{|2\alpha^3 - 2\alpha^2\gamma|}{|\alpha - \gamma|} + |\beta^2 - 1| + |\beta + 1|$ Since $\alpha < \gamma$, $|\alpha - \gamma| = \gamma - \alpha$ and factor numerator: $|2\alpha^2(\alpha - \gamma)| = 2\alpha^2 |\alpha - \gamma| = 2\alpha^2(\gamma - \alpha)$ So first term is $\frac{2\alpha^2(\gamma - \alpha)}{\gamma - \alpha} = 2\alpha^2$ Thus $\Delta = 2\alpha^2 + |\beta^2 - 1| + |\beta + 1|$ 5. For $0 < x < y < 3$: a) $A = d(3, x) - |y - 3|$ where $d(a,b) = |a - b|$ $d(3,x) = |3 - x| = 3 - x$ since $x < 3$ Also $|y - 3| = 3 - y$ since $y < 3$ So $A = (3 - x) - (3 - y) = y - x$ b) $B = 3|x - 5| + 2|-y + x| - 5|x + y - 6|$ Since $x < y < 3 < 5$, $x-5 < 0$, so $|x-5|=5 - x$ Also $-y + x = x - y < 0$ because $x < y$, so $|-y + x|=y - x$ $x + y - 6 < 3 + 3 - 6 = 0$, so $|x + y - 6|=6 - x - y$ Calculate: $3(5 - x) + 2(y - x) - 5(6 - x - y) = 15 - 3x + 2y - 2x - 30 + 5x + 5y = (15 - 30) + (-3x - 2x + 5x) + (2y + 5y) = -15 + 0 + 7y = 7y - 15$ c) $\Gamma = d(x^2 + 2x, -1) - | -x^2 - 1|$ First term: $|x^2 + 2x + 1|$ since distance to -1 is $|value + 1|$ Note $x^2 + 2x + 1 = (x + 1)^2 \ge 0$, so $|x^2 + 2x + 1| = (x + 1)^2$ Second term: $|-x^2 - 1| = |-(x^2 + 1)| = x^2 + 1$ So $\Gamma = (x + 1)^2 - (x^2 + 1) = x^2 + 2x + 1 - x^2 - 1 = 2x$ d) $\Delta = \frac{x^2 + 8|x| + 16}{|x| + 4} + \frac{y^2 + |y|}{|y|}$ Since $x > 0$, $|x| = x$, similarly $|y| = y$ Numerators: $x^2 + 8x + 16$ and $y^2 + y$ Denominators: $x + 4$ and $y$ So $\Delta = \frac{x^2 + 8x + 16}{x + 4} + \frac{y^2 + y}{y} = \frac{(x + 4)^2}{x + 4} + y + 1 = (x + 4) + y + 1 = x + y + 5$ 6. Simplifications: a) $A = 1 - |x^2 + 1|$ Since $x^2 + 1 > 0$ always, $|x^2 + 1| = x^2 + 1$ So $A = 1 - (x^2 + 1) = -x^2$ b) (Labeled c) $\Gamma = |x - 1| + |2 - x|$ For $x$ between 1 and 2, both positive: $|x-1|=x-1$, $|2 - x|=2 - x$ Sum is $1$ c) $E = |2x - 4| + |2 - x| - |5x - 10|$ Note $|2x - 4| = 2|x - 2|$, $|5x - 10| = 5|x - 2|$ So expression is $2|x - 2| + |2 - x| - 5|x - 2|$ But $|2 - x|=|x - 2|$, so sum is: $2|x - 2| + |x - 2| - 5|x - 2| = (2 + 1 - 5)|x - 2| = -2|x - 2|$ d) $B = 3 + |x + 3| + x$ No simplification unless domain specified. e) $\Delta = |x^2 + 4x + 6| - x^2 - 2x - 2$ Inside absolute value, quadratic discriminant negative, $(x^2 + 4x + 6) > 0$ always. So $\Delta = x^2 + 4x + 6 - x^2 - 2x - 2 = 2x + 4$ f) $\Sigma = |2x - 4| - |3 - 3x| + |x^2 + 21|$ The last term $|x^2 + 21| = x^2 + 21$ always positive. Simplifying depends on $x$, so expression remains as above. Final answers and simplifications for each part shown stepwise.