جواب تمرین صفحه 93 درس 5 ریاضی نهم (عبارت های جبری)

الف \(a - b = 1 \to a = b + 1 \Rightarrow a > b\)

ب \(u - v = - 2 \to u = v - 2 \Rightarrow u < v\)

ج

\(\begin{array}{l}2p - 2 = 2q - 3 \to 2p = 2q - 1\,\,\,\,\,\mathop \to \limits^{ \times \frac{1}{2}} \\\\p = q - \frac{1}{2} \Rightarrow p < q\end{array}\)

د

\(\begin{array}{l}\frac{{a - b}}{2} = - 3\,\,\,\,\,\mathop \to \limits^{ \times 2} \,\,\,\,\,a - b = - 6 \to \\\\a = b - 6 \Rightarrow a < b\end{array}\)

الف \(\frac{{{a^2}}}{{bc}} > 0\)

حالت اول: \(a > 0\,,\,c < 0 \to \frac{{ac}}{{{b^2}}} < 0\)

حالت دوم: \(a < 0\,,\,c > 0 \to \frac{{ac}}{{{b^2}}} < 0\)

ب \(\frac{a}{{bc}} > 0\)

حالت اول: \(a > 0\,,\,b < 0\,,\,c < 0\)

حالت دوم: \(a < 0\,,\,b < 0\,,\,c > 0\)

حالت سوم: \(a < 0\,,\,b > 0\,,\,c < 0\)

حالت چهارم: \(a > 0\,,\,b > 0\,,\,c > 0\)

ج \(ab > 0\)

حالت اول: \(a > 0\,,\,b > 0\,\)

حالت دوم: \(a < 0\,,\,b < 0\,\)

د \(\frac{{{a^2}}}{{bc}} > 0\)

حالت اول: \(b > 0\,,\,c > 0\,\)

حالت دوم: \(b < 0\,,\,c > 0\,\)

الف

\(\begin{array}{l}2x - 6 + 5 < 5 - x \to 2x - 1 < 5 - x\\\\ \to 3x < 6 \to x < 2\\\\D = \{ x \in \mathbb{R}|x < 2\} \end{array}\)

ب

\(\begin{array}{l}3 - 2x \ge 15 - 10x \to - 2x + 10x \ge 15 - 3\\\\ \to 8x \ge 12 \to x\frac{{12}}{8} \to x \ge \frac{3}{2}\\\\D = \{ x \in \mathbb{R}|x \le \frac{3}{2}\} \end{array}\)

ج

\(\begin{array}{l}\frac{{y - 3 - 4}}{4} < \frac{y}{2}\,\,\,\,\,\mathop \to \limits^{ \times 4} \,\,\,\,\,y - 7 < 2y\\\\ \to y - 2y < 7 \to - y > 7 \to y < - 7\\\\D = \{ x \in \mathbb{R}|y \le 7\} \end{array}\)

د

\(\begin{array}{l}\frac{{ - 8 - q}}{4} \le \frac{{1 + q}}{3}\,\,\,\,\,\mathop \to \limits^{ \times 12} \,\,\,\,\, - 24 - 3q \le 4 + 4q\\\\ \to - 4q - 3q \le 4 + 24\\\\ \to - 7q \le 28 \to q \ge \frac{{ - 28}}{7} \to q \ge - 4\\\\D = \{ x \in \mathbb{R}|q \le - 4\} \end{array}\)

خیر

\(\begin{array}{l}\left\{ \begin{array}{l}a,b > 0\\{a^2} > {b^2}\end{array} \right. \to {a^2} - {b^2} > 0 \to \\\\(a - b)(a + b) > 0 \to \\\\\left\{ \begin{array}{l}(a - b) > 0,\,(a + b) > 0 \to \left\{ \begin{array}{l}a - b > 0 \to a > b\\a + b > 0 \to a > - b\end{array} \right.\\(a - b) < 0,\,(a + b) < 0 \to \end{array} \right.\end{array}\)

مورد قبول نیست چون a,b>0 هستند

الف \(3x \ge 2x + 300\)

ب \(\frac{a}{2} + 4b \le 6\)

مقدار پول مورد نیاز هر روز

\(\begin{array}{l}(3 \times 65) + (3 \times 85) = 450\\\\450x \le 4500\\\\x \le 10\end{array}\)