星期四��上21点，将会播出一集女����的����。这个����将会������都市��拉利�����������和����。女��们��供��在����上，星期四��上21点，将会播出一集女����的����。这个����将会������都市��拉利�����������和����。女��们��供��在����上，其中也包����称为“美之女��”的����。��尔·������从���������生还后，本应该享��这场��会。然而，直到他收到一��来自��位����������的信，他��意��到这次女����可能会有些不同��常。信中提到，��尔��和他���自约会。在����都市的��个角��，少女��注一��地��出��定，这使得��尔和整个����都市都��入了������之中。与此同时，号称“最��”的“���������们”也开始行动了。这将是少年的���步，少女的期�� - 【��族之物语】。<|endoftext|> \begin{tikzpicture} % Define nodes \node[obs] (x) {$\mathbf{x}$}; \node[latent, above=of x, xshift=-0.5cm] (z) {$\mathbf{z}$}; \node[latent, above=of x, xshift=0.5cm] (theta) {$\boldsymbol{\theta}$}; \node[latent, below=of x, xshift=0.75cm] (w) {$\mathbf{w}$}; % Connect the nodes \edge {z} {x} \edge {theta} {x} \edge {w} {x} % Plates \plate {plate1} {(x)(z)(theta)} {$N$}; \plate {plate2} {(x)(w)} {$M$}; \end{tikzpicture}<|endoftext|>Varje tal som är summan av två perfekta kvadrater är delbar med fyra.<|endoftext|>## Problem 1Evaluate the following numerical expressions.a) $3 4$b) $7-2$c) $5-8$d) $12 10$a) $3 4 = 7$b) $7-2 = 5$c) $5-8 = -3$d) $12 10 = 22$<|endoftext|>Algebraic<|endoftext|>#define ID 1#define NAME "John Doe"#define AGE 30#define PI 3.14159265359<|endoftext|>## Problem 1Let $a, b, c$ be positive real numbers such that $a b c=1$. Prove that$$\frac{a}{a^2 1} \frac{b}{b^2 1} \frac{c}{c^2 1} \leq \frac{9}{10}.$$Solution:We begin by noting that for any positive real number $x$, we have$$x^2 1 \geq 2x,$$since this inequality is equivalent to$$x^2-2x 1=(x-1)^2 \geq 0,$$which is clearly true. Hence, for any positive real numbers $a, b, c$, we have$$\frac{a}{a^2 1} \leq \frac{a}{2a} = \frac{1}{2}.$$Similarly, we have $\frac{b}{b^2 1} \leq \frac{1}{2}$ and $\frac{c}{c^2 1} \leq \frac{1}{2}$. Summing these three inequalities, we obtain$$\frac{a}{a^2 1} \frac{b}{b^2 1} \frac{c}{c^2 1} \leq \frac{1}{2} \frac{1}{2} \frac{1}{2} = \frac{3}{2}.$$However, since $a b c=1$, we have $a, b, c < 1$, and hence $a^2, b^2, c^2 < 1$. This implies that $a^2 1, b^2 1, c^2 1 < 2$, and therefore, $\frac{1}{a^2 1}, \frac{1}{b^2 1}, \frac{1}{c^2 1} > \frac{1}{2}$. Multiplying these three inequalities, we obtain$$\frac{1}{(a^2 1)(b^2 1)(c^2 1)} > \frac{1}{2^3} = \frac{1}{8}.$$Raising both sides of this inequality to the $3$rd power, we obtain$$\frac{1}{(a^2 1)^3(b^2 1)^3(c^2 1)^3} > \frac{1}{8^3} = \frac{1}{512}.$$Multiplying both sides of this inequality by $a^3b^3c^3$, we obtain$$\frac{a^3b^3c^3}{(a^2 1)^3(b^2 1)^3(c^2 1)^3} > \frac{a^3b^3c^3}{512}.$$However, by the AM-GM inequality, we have$$\frac{a^3b^3c^3}{512} \leq \left(\frac{a b c}{3}\right)^9 = \frac{1}{3^9} = \frac{1}{19683}.$$Combining this inequality with the previous one, we obtain$$\frac{a^3b^3c^3}{(a^2 1)^3(b^2 1)^3(c^2 1)^3} > \frac{1}{19683}.$$Finally, taking reciprocals on both sides of this inequality, we obtain$$(a^2 1)^3(b^2 1)^3(c^2 1)^3 > 19683,$$which implies that$$(a^2 1)(b^2 1)(c^2 1) > 27.$$Hence, we have$$\frac{1}{a^2 1} \cdot \frac{1}{b^2 1} \cdot \frac{1}{c^2 1} < \frac{1}{27}.$$Multiplying both sides of this inequality by $a^2b^2c^2$, we obtain$$\frac{a^2b^2c^2}{(a^2 1)(b^2 1)(c^2 1)} < \frac{a^2b^2c^2}{27}.$$However, by the AM-GM inequality, we have$$\frac{a^2b^2c^2}{27} \leq \left(\frac{a b c}{3}\right)^6 = \frac{1}{3^6} = \frac{1}{729}.$$Combining this inequality with the previous one, we obtain$$\frac{a^2b^2c^2}{(a^2 1)(b^2 1)(c^2 1)} < \frac{1}{729},$$and hence,$$\frac{a^2}{a^2 1} \cdot \frac{b^2}{b^2 1} \cdot \frac{c^2}{c^2 1} < \frac{1}{729}.$$Multiplying both sides of this inequality by $a^2 b^2 c^2$, we obtain$$\frac{a^2}{a^2 1} \cdot \frac{b^2}{b^2 1} \cdot \frac{c^2}{c^2 1} \cdot (a^2 b^2 c^2) < \frac{a^2 b^2 c^2}{729}.$$However, by the Cauchy-Schwarz inequality, we have$$(a^2 b^2 c^2)\left(\frac{a^2}{a^2 1} \frac{b^2}{b^2 1} \frac{c^2}{c^2 1}\right) \geq (a b c)^2 = 1,$$and hence,$$\frac{a^2}{a^2 1} \frac{b^2}{b^2 1} \frac{c^2}{c^2 1} \geq \frac{1}{a^2 b^2 c^2} > 1.$$Combining this inequality with the previous one, we obtain$$\frac{a^2}{a^2 1} \cdot \frac{b^2}{b^2 1} \cdot \frac{c^2}{c^2 1} \cdot (a^2 b^2 c^2) < \frac{1}{729} \cdot \frac{a^2 b^2 c^2}{a^2 b^2 c^2} = \frac{1}{729}.$$Finally, multiplying both sides of this inequality by $9$, we obtain$$\frac{9a^2}{a^2 1} \cdot \frac{9b^2}{b^2 1} \cdot \frac{9c^2}{c^2 1} < \frac{9}{729},$$which simplifies to$$\frac{a^2}{a^2 1} \cdot \frac{b^2}{b^2 1} \cdot \frac{c^2}{c^2 1} < \frac{1}{81}.$$Combining this inequality with the inequality we obtained earlier, we obtain$$\frac{a^2}{a^2 1} \cdot \frac{b^2}{b^2 1} \cdot \frac{c^2}{c^2 1} < \frac{1}{81} < \frac{1}{27} < \frac{1}{8} < \frac{1}{2} < \frac{3}{2}.$$Therefore, we have$$\frac{a^2}{a^2 1} \frac{b^2}{b^2 1} \frac{c^2}{c^2 1} < \frac{3}{2}.$$Combining this inequality with the inequalities we obtained at the beginning of the solution, we obtain$$\frac{a}{a^2 1} \frac{b}{b^2 1} \frac{c}{c^2 1} < \frac{1}{2} \frac{1}{2} \frac{1}{2} = \frac{3}{2} < \frac{9}{10}.$$Hence, we have$$\frac{a}{a^2 1} \frac{b}{b^2 1} \frac{c}{c^2 1} < \frac{9}{10},$$as desired. $\square$<|endoftext|># Language: Python 3 Notebook# Language: Python#1. 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