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حل المعادلات التالية باستعمال الخواص الأعداد الحقيقية وتحقق من صحة الحل:

$z = \sqrt[3]{- 8} + 2 z$

$z = \sqrt[3]{- 8} + 2 z \Rightarrow z = - 2 + 2 z \Rightarrow 2 = 2 z - z \Rightarrow z = 2 2 = \sqrt[3]{- 8} + 2 (2) \Rightarrow 2 = - 2 + 4 \Rightarrow 2 = 2$

$4 m - 5 \sqrt{3} = 3 m + 6 \sqrt{3}$

$4 m - 5 \sqrt{2} = 3 m + 6 \sqrt{2} \Rightarrow 4 m - 3 m = 5 \sqrt{2} + 6 \sqrt{2} \Rightarrow m = 11 \sqrt{2}$

$\sqrt{36} h = | - 16 | + 5 h$

$\sqrt{36} h = | - 16 | + 5 h \Rightarrow 6 h = 16 + 5 h \Rightarrow h = 16$

$3 z \div 21 = \frac{1}{7}$

$3 z \div 21 = \frac{1}{7} \Rightarrow \frac{3 z}{21} = \frac{1}{7} \Rightarrow \frac{1}{7} z = \frac{1}{7} \Rightarrow z = 1$

$\sqrt[3]{125} \times \div | - 9 | = 5^{2}$

$\begin{aligned} \sqrt[3]{125} x \div | - 19 | = 5^{2} \Rightarrow 5 x \div 19 = 25 \Rightarrow \frac{5 x}{19} = 25 \\ \Rightarrow 5 x = 225 \Rightarrow x = \frac{225}{5} \Rightarrow x = 45 \end{aligned}$

$\sqrt{3} x \div 9 = \sqrt{3} \div 3$

$\sqrt{3} x \div 9 = \sqrt{3} \div 3 \Rightarrow \frac{\sqrt{3} x}{9} = \frac{\sqrt{3}}{3} \Rightarrow \frac{9}{\sqrt{3}} (\frac{\sqrt{3 x}}{9}) = \frac{9}{\sqrt{3}} \times \frac{\sqrt{3}}{3} \Rightarrow x = 3$

$\frac{\sqrt{5} h}{2} = \frac{1}{\sqrt{5}}$

$\frac{\sqrt{5}}{2} h = \frac{1}{\sqrt{5}} \Rightarrow \sqrt{5} \cdot \sqrt{5} h = 2 \Rightarrow 5 h = 2 \Rightarrow h = \frac{2}{5}$

$\frac{9 y}{2 \sqrt[3]{4}} = \frac{18}{4}$

$\frac{9 y}{3 \sqrt[3]{4}} = \frac{18}{4} \Rightarrow 36 y = 54 \sqrt[3]{4} \Rightarrow y = \frac{54 \sqrt[3]{4}}{36} \Rightarrow y = \frac{9 \sqrt[3]{4}}{4}$

$6 z \div 13 = 5 z \div 13$

$6 z \div 13 = 2 z \div 13 \Rightarrow \frac{6 z}{13} = \frac{2 z}{13} \Rightarrow 6 z = 2 z \Rightarrow z = 0$

$8 (h - 1^{2}) = \frac{1}{2} h - 6$

$\begin{aligned} 8 (h - 1^{2}) = \frac{1}{2} h - 6 \Rightarrow 8 h - 8 = \frac{1}{2} h - 6 \Rightarrow 8 h - \frac{1}{2} h = 8 - 6 \\ \Rightarrow \frac{15}{2} h = 2 \Rightarrow h = \frac{4}{15} \end{aligned}$

$5 \sqrt{3} - z = z - 7 \sqrt{3}$

$\begin{aligned} 5 \sqrt{3} - z = z - 7 \sqrt{3} \Rightarrow - z - z = - 5 \sqrt{3} - 7 \sqrt{3} \Rightarrow - 2 z = - 12 \sqrt{3} \\ \Rightarrow z = 6 \sqrt{3} \end{aligned}$

$\sqrt{64} y = 10 (y - 1) + 3^{2}$

$\begin{aligned} \sqrt{64} y = 10 (y - 1) + 3^{2} \Rightarrow 8 y = 10 y - 10 + 9 \Rightarrow 8 y - 10 y = - 1 \\ \Rightarrow - 2 y = - 1 \Rightarrow y = \frac{1}{2} \end{aligned}$

حل المعادلات التالية في R باستعمال الخواص:

$3 (x - 10) = 2 (x + 10)$

$\begin{aligned} 3 (x - 10) = 2 (x + 10) \Rightarrow 3 x - 30 = 2 x + 20 \\ \Rightarrow 3 x - 2 x = 30 + 20 \Rightarrow x = 50 \end{aligned}$

$\sqrt[3]{- 8} y \div | - 8 | = 4 \sqrt{2}$

$\begin{aligned} \sqrt[3]{- 8} y \div | - 8 | = 4 \sqrt{2} \Rightarrow \frac{- 2 y}{8} = 4 \sqrt{2} \Rightarrow - 2 y = 32 \sqrt{2} \\ \Rightarrow y = \frac{32 \sqrt{2}}{- 2} \Rightarrow y = - 16 \sqrt{2} \end{aligned}$

$\frac{1}{3} (z - 7) + \frac{7}{3} = \frac{1}{5} (z - 10)$

$\begin{aligned} \frac{1}{3} (z - 7) + \frac{7}{3} = \frac{1}{5} (z - 10) \Rightarrow \frac{1}{3} z - \frac{7}{3} + \frac{7}{3} = \frac{1}{5} z - 2 \\ \Rightarrow \frac{1}{3} z - \frac{1}{5} z = - 2 \Rightarrow \frac{5 - 3}{15} z = - 2 \Rightarrow 2 z = - 30 \Rightarrow z = - 15 \end{aligned}$

$\frac{t}{8 + \sqrt[3]{- 27}} = \frac{6 t}{5}$

$\frac{t}{8 + \sqrt[3]{- 27}} = \frac{6 t}{5} \Rightarrow \frac{t}{8 - 3} = \frac{6 t}{5} \Rightarrow \frac{t}{5} = \frac{6 t}{5} \Rightarrow t = 6 t \Rightarrow t = 0$

$| y - 12 | = 7$

$| 2 y - 12 | = 7 \Rightarrow {\begin{cases} y - 12 = 7 \Rightarrow y = 19 \\ y - 12 = - 7 \Rightarrow y = 5 \end{cases}$

مجموعة الحل هي: ${19, 5}$

$| 2 v - 5 | = \sqrt{36}$

$| 2 v - 5 | = \sqrt{36} \Rightarrow {\begin{cases} 2 v - 5 = 6 \Rightarrow 2 v = 11 \Rightarrow v = \frac{11}{2} \\ 2 v - 5 = - 6 \Rightarrow 2 v = - 1 \Rightarrow v = \frac{- 11}{2} \end{cases}$

مجموعة الحل هي: ${\frac{11}{2}, \frac{- 1}{2}}$

$| \frac{1}{3} n + 8 | = \sqrt[3]{- 125}$

$| \frac{1}{3} n + 18 | = \sqrt[3]{- 125} \Rightarrow | \frac{1}{3} n + 18 | = - 5$

لا يمكن لأن المسافة سالبة فمجموعة الحل هي: $\emptyset$

$| 7 x - 14 | = | - 18 |$

$\Rightarrow | 7 x - 14 | = 18 \Rightarrow {\begin{cases} 7 x - 14 = 18 \Rightarrow 7 x = 32 \Rightarrow x = \frac{32}{7} \\ 7 x - 14 = - 18 \Rightarrow 2 x = - 4 \Rightarrow x = \frac{- 4}{7} \end{cases}$

مجموعة الحل هي: ${\frac{32}{7}, \frac{- 4}{7}}$

$| z - \sqrt{3} | = 4 \sqrt{3}$

$| z - \sqrt{3} | = 4 \sqrt{3} \Rightarrow {\begin{cases} z - \sqrt{3} = 4 \sqrt{3} \Rightarrow z = 5 \sqrt{3} \\ z - \sqrt{3} = - 4 \sqrt{3} \Rightarrow z = - 3 \sqrt{3} \end{cases}$

مجموعة الحل هي: ${5 \sqrt{3}, - 3 \sqrt{3}}$

حل المعادلات التالية في R باستعمال الجذر التربيعي:

$x^{2} = 64$

$x^{2} = 64 \Rightarrow x = 8 o r x = - 8$

$9 y^{2} = 1$

$9 x^{2} = 1 \Rightarrow x^{2} = \frac{1}{9} \Rightarrow x = \frac{1}{3} o r x = - \frac{1}{3}$

$12 t^{2} = 4$

$12 t^{2} = 4 \Rightarrow t^{2} = \frac{4}{12} \Rightarrow t = \frac{2}{2 \sqrt{3}} o r t = - \frac{2}{2 \sqrt{3}} \Rightarrow t = \frac{1}{\sqrt{3}} o r t = - \frac{1}{\sqrt{3}}$

$n^{2} - 5 = 20$

$n^{2} - 5 = 20 \Rightarrow n^{2} = 25 \Rightarrow n = 5 o r n = - 5$

$7 - z^{2} = - 42$

$7 - z^{2} = - 42 \Rightarrow - z^{2} = - 42 - 7 \Rightarrow z^{2} = 49 \Rightarrow z = 7 o r z = - 7$

$\frac{1}{4} k^{2} = 9$

$\frac{1}{4} k^{2} = 9 \Rightarrow k^{2} = 36 \Rightarrow k = 6 o r k = - 6$

$y^{2} = \frac{4}{9}$

$y^{2} = \frac{4}{9} \Rightarrow y = \frac{2}{3} o r y = - \frac{2}{3}$

$h^{2} - \frac{3}{4} = \frac{1}{4}$

$h^{2} - \frac{3}{4} = \frac{1}{4} \Rightarrow h^{2} = \frac{1 + 3}{4} \Rightarrow h^{2} = \frac{4}{4} \Rightarrow h^{2} = 1 \Rightarrow h = 1 o r h = - 1$

حل المعادلات التالية في R باستعمال خاصية الضرب الصفري:

$(y - 4) (y + 4) = 0$

$\begin{aligned} (y - 4) (y + 4) = 0 \Rightarrow (y - 4) = 0 o r (y + 4) = 0 \\ \Rightarrow y = 4 o r y = - 4 \end{aligned}$

$(z - 7) (z - 7) = 0$

$(z - 7) (z - 7) = 0 \Rightarrow z - 7 = 0 o r z - 7 = 0 \Rightarrow z = 7$

$(x + \sqrt{5}) (x - \sqrt{3}) = 0$

$\begin{aligned} (x + \sqrt{5}) (x - \sqrt{3}) = 0 \Rightarrow x + \sqrt{5} = 0 o r x - \sqrt{3} = 0 \\ \Rightarrow x = - \sqrt{5} o r x = \sqrt{3} \end{aligned}$

$(\sqrt{2} - h) (\sqrt{2} + h) = 0$

$\begin{aligned} (\sqrt{2} - h) (\sqrt{2} + h) = 0 \Rightarrow \sqrt{2} - h = 0 o r \sqrt{2} + h = 0 \\ \Rightarrow h = \sqrt{2} o r h = - \sqrt{2} \end{aligned}$

$(4 t + 8) (3 t - 7) = 0$

$\begin{aligned} (4 t + 8) (3 t - 7) = 0 \Rightarrow 4 t + 8 = 0 o r 3 t - 7 = 0 \\ t = - 2 o r t = \frac{7}{3} \end{aligned}$

$z^{2} - z = 0$

$z^{2} - z = 0 \Rightarrow z (z - 1) = 0 \Rightarrow z = 0 o r z = 1$

$\sqrt{8} x^{2} + 2 x = 0$

$\sqrt{8} x^{2} + 2 x = 0 \Rightarrow 2 \sqrt{2} x^{2} + 2 x = 0 \Rightarrow 2 x (\sqrt{2} x + 1) = 0 \Rightarrow 2 x = 0 o r \sqrt{2} x + 1 = 0 \Rightarrow x = 0 o r x = \frac{- 1}{\sqrt{2}}$

$3 \sqrt{7} n^{2} - 3 \sqrt{7} n = 0$

$\begin{aligned} 3 \sqrt{7} n^{2} - 3 \sqrt{7} n = 0 \Rightarrow 3 \sqrt{7} n (n - 1) = 0 \Rightarrow 3 \sqrt{7} n = 0 o r n - 1 = 0 \\ \Rightarrow n = 0 o r n = 1 \end{aligned}$

$\frac{1}{5} y^{2} - \frac{1}{5} y = 0$

$\begin{aligned} 3 \frac{1}{5} y^{2} - \frac{1}{5} y = 0 \Rightarrow \frac{1}{5} y (y - 1) = 0 \Rightarrow \frac{1}{5} y = 0 o r y - 1 = 0 \\ \Rightarrow y = 0 o r y = 1 \end{aligned}$

حل المتباينات التالية باستعمال خواص المتباينات على الأعداد الحقيقية:

$2 (x - \sqrt{2}) \geq x - \sqrt{2}$

$\begin{aligned} 2 (x - \sqrt{2}) \geq x - \sqrt{2} \Rightarrow 2 x - \sqrt{2} \geq x - \sqrt{2} \\ \Rightarrow 2 x - x \geq 2 \sqrt{2} - \sqrt{2} \Rightarrow\geq x \geq \sqrt{2} \end{aligned}$

$9 z + \sqrt[3]{- 27} < 10 z - \sqrt[3]{125}$

$\begin{aligned} 9 z + \sqrt[3]{- 27} < 10 z - \sqrt[3]{125} \Rightarrow 9 z - 3 < 10 z - 5 \\ \Rightarrow 9 z - 10 z < 3 - 5 \Rightarrow - z < - 2 \Rightarrow z > 2 \end{aligned}$

$5 (\frac{1}{2} y - \frac{3}{10}) \leq 0$

$5 (\frac{1}{2} y - \frac{3}{10}) \leq 0 \Rightarrow \frac{5}{2} y - \frac{3}{2} \leq 0 \Rightarrow \frac{5}{2} y \leq \frac{3}{2} \Rightarrow 5 y \leq 3 \Rightarrow y \leq \frac{3}{5}$

$\frac{t}{- 7} - 1 > \frac{- 1}{14}$

$\begin{aligned} \frac{t}{- 7} - 1 > \frac{- 1}{14} \Rightarrow \frac{t}{- 7} > \frac{- 1}{14} + 1 \Rightarrow \frac{t}{- 7} > \frac{- 1 + 14}{14} \\ \Rightarrow \frac{t}{- 7} > \frac{13}{14} \Rightarrow - t > \frac{13}{2} \Rightarrow t < \frac{- 13}{2} \end{aligned}$

$\frac{2 m}{9} - \frac{1}{9} \leq \frac{1}{9}$

$\frac{2 m}{9} - \frac{1}{9} \leq \frac{1}{9} \Rightarrow \frac{2 m}{9} \leq \frac{1}{9} + \frac{1}{9} \Rightarrow \frac{2 m}{9} \leq \frac{2}{9} \Rightarrow 2 m \leq 2 \Rightarrow m \leq 1$

$3 (x + 7) < 6 - \sqrt[3]{- 64}$

$\begin{aligned} 3 (x + 7) < 6 - \sqrt[3]{- 64} \Rightarrow 3 x + 21 < 6 + 4 \Rightarrow 3 x < 10 - 21 \\ \Rightarrow 3 x < - 11 \Rightarrow x < \frac{- 11}{3} \end{aligned}$

$\frac{1}{6} (z - \frac{12}{5}) + \frac{5}{6} z \geq - \frac{3}{5}$

$\frac{1}{6} (z - \frac{12}{5}) + \frac{5}{6} z \geq \frac{- 3}{5} \Rightarrow \frac{1}{6} z + \frac{5}{6} z - \frac{2}{5} \geq \frac{- 3}{5} \Rightarrow \frac{1 + 5}{6} z \geq \frac{- 3}{5} + \frac{2}{5} \Rightarrow \frac{6}{6} z \geq \frac{- 3 + 2}{5} \Rightarrow z \geq \frac{- 1}{5}$

$\frac{y}{4} \leq 2 (\frac{1}{\sqrt{16}} - \frac{1}{8} y)$

$\frac{y}{4} \leq 2 (\frac{1}{\sqrt{16}} - \frac{1}{8} y) \Rightarrow \frac{y}{4} \leq \frac{2}{4} - \frac{1}{4} y \Rightarrow \frac{y}{4} + \frac{y}{4} \leq \frac{1}{2} \Rightarrow \frac{1}{2} y \leq \frac{1}{2} \Rightarrow y \leq 1$

$5 (x + 1) > 2 (1 - x)$

$5 (x + 1) > 2 (1 - x) \Rightarrow 5 x + 5 > 2 - 2 x \Rightarrow 5 x + 2 x > 2 - 5 \Rightarrow 7 x > - 3 \Rightarrow x > \frac{- 3}{7}$

$\frac{1}{3} (h - \sqrt{2}) \leq \frac{2}{3} (\sqrt{2} - h)$

$\frac{1}{3} (h - \sqrt{2}) \leq \frac{2}{3} (\sqrt{2} - h) \Rightarrow \frac{1}{3} h - \frac{\sqrt{2}}{3} \leq \frac{2 \sqrt{2}}{3} - \frac{2}{3} h \Rightarrow \frac{1}{3} h + \frac{2}{3} h \leq \frac{2 \sqrt{2}}{3} + \frac{\sqrt{2}}{3} \Rightarrow \frac{3}{3} h \leq \frac{3 \sqrt{2}}{3} \Rightarrow h \leq \sqrt{2}$

$\frac{m}{6} + \frac{2}{5} < \frac{m}{3} - \frac{4}{5}$

$\frac{m}{6} + \frac{2}{5} < \frac{m}{3} - \frac{4}{5} \Rightarrow \frac{m}{6} - \frac{m}{3} < - \frac{4}{5} - \frac{2}{5} \Rightarrow \frac{1 - 2}{6} m < \frac{- 4 - 2}{5} \Rightarrow \frac{- 1}{6} m < \frac{- 6}{5} \Rightarrow - m < \frac{- 36}{5} \Rightarrow m > \frac{36}{5}$

$\frac{9}{\sqrt[3]{- 27}} - 5 y > \frac{1}{5} - y$

$\frac{9}{\sqrt[3]{- 27}} - 5 y > \frac{1}{5} - y \Rightarrow \frac{9}{- 3} - 5 y > \frac{1}{5} - y \Rightarrow - 3 - 5 y > \frac{1}{5} - y \Rightarrow - 5 y + y > \frac{1}{5} + 3 \Rightarrow - 4 y > \frac{1 + 15}{5} \Rightarrow - 4 y > \frac{16}{5} \Rightarrow y < \frac{- 16}{20} \Rightarrow y < \frac{- 4}{5}$

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حمل تطبيق دراستي من متجر جوجل

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