حلول الأسئلة

السؤال

اختر الإجابة الصحيحة: حل المتباينات التالية في R باستعمال خواص المتباينات على الأعداد الحقيقية:

الحل

$\begin{aligned} \frac{- 7}{8} (\frac{8}{14} x + \frac{1}{\sqrt{7}}) \geq 0 \end{aligned}$

$\begin{aligned} x \leq \frac{4}{\sqrt{7}} \end{aligned}$
$\begin{aligned} x \leq \frac{\sqrt{7}}{4} \end{aligned}$
$\begin{aligned} x \geq \frac{4}{\sqrt{7}} \end{aligned}$
$\begin{aligned} x \leq \frac{- \sqrt{7}}{4} \end{aligned}$

مشاركة الحل

فايسبوك

واتساب

تيليجرام

طباعة

[4-5] حل المتباينات الجبرية متعددة الخطوات في R

اختر الإجابة الصحيحة لكل مما يأتي:

حل المتباينات التالية في R باستعمال خواص المتباينات على الأعداد الحقيقية:

$5 (y + 1) \geq 8 - \sqrt[3]{- 125}$

$\begin{aligned} s = {y : y \in R, y \leq - \frac{5}{8}} \end{aligned}$
$\begin{aligned} s = {y : y \in R, y \leq \frac{8}{5}} \end{aligned}$
$\begin{aligned} s = {y : y \in R, y \geq \frac{5}{8}} \end{aligned}$
$\begin{aligned} s = {y : y \in R, y \geq \frac{8}{5}} \end{aligned}$

$\begin{aligned} \frac{1}{2} (x - \frac{4}{3}) + \frac{1}{6} x < - \frac{2}{3} \end{aligned}$

$x < \frac{4}{3}$
$x > 0$
$x < 0$
$x > \frac{- 4}{3}$

$\begin{aligned} \frac{z}{6} \leq 4 (\frac{1}{\sqrt{9}} - \frac{1}{3} z) \end{aligned}$

$z \leq - 4$
$z \geq - 4$
$z \geq 4$
$z \leq \frac{8}{9}$

$\begin{aligned} 2 (v - 4 \sqrt{5}) > 5 (v - \sqrt{5}) \end{aligned}$

$\begin{aligned} v < - \sqrt{5} \end{aligned}$
$\begin{aligned} v > \sqrt{5} \end{aligned}$
$\begin{aligned} v > - \sqrt{5} \end{aligned}$
$\begin{aligned} v < \sqrt{5} \end{aligned}$

$\begin{aligned} \frac{3}{2} n - \sqrt[3]{- 8} - \frac{1}{4} n \leq | - 5 | \end{aligned}$

$\begin{aligned} n < \frac{12}{5} \end{aligned}$
$\begin{aligned} n \leq \frac{12}{5} \end{aligned}$
$\begin{aligned} n \leq \frac{- 12}{5} \end{aligned}$
$\begin{aligned} n < \frac{- 12}{5} \end{aligned}$

$\begin{aligned} 12 h - \sqrt[3]{125} > 6 (h + 1) \end{aligned}$

$\begin{aligned} h > \frac{6}{11} \end{aligned}$
$\begin{aligned} h < \frac{11}{6} \end{aligned}$
$\begin{aligned} h > \frac{11}{6} \end{aligned}$
$\begin{aligned} h < \frac{6}{11} \end{aligned}$

$\begin{aligned} \frac{- 7}{8} (\frac{8}{14} x + \frac{1}{\sqrt{7}}) \geq 0 \end{aligned}$

$\begin{aligned} x \leq \frac{4}{\sqrt{7}} \end{aligned}$
$\begin{aligned} x \leq \frac{\sqrt{7}}{4} \end{aligned}$
$\begin{aligned} x \geq \frac{4}{\sqrt{7}} \end{aligned}$
$\begin{aligned} x \leq \frac{- \sqrt{7}}{4} \end{aligned}$

$\begin{aligned} 11 (2 - y) < 4 (5 + y) \end{aligned}$

$\begin{aligned} y > \frac{15}{2} \end{aligned}$
$\begin{aligned} y < \frac{2}{15} \end{aligned}$
$\begin{aligned} y > \frac{- 2}{15} \end{aligned}$
$\begin{aligned} y > \frac{2}{15} \end{aligned}$

$\begin{aligned} \frac{1}{5} (z - \sqrt{3}) \geq \frac{1}{5} (\sqrt{3} - z) \end{aligned}$

$\begin{aligned} z \geq \sqrt{3} \end{aligned}$
$\begin{aligned} z \leq \frac{2}{\sqrt{3}} \end{aligned}$
$\begin{aligned} z \leq \sqrt{3} \end{aligned}$
$\begin{aligned} z \geq \frac{2}{\sqrt{3}} \end{aligned}$

$\begin{aligned} \frac{t}{2} - \frac{1}{7} < \frac{t}{14} + \frac{5}{14} \end{aligned}$

$\begin{aligned} t > \frac{6}{7} \end{aligned}$
$\begin{aligned} t < \frac{7}{6} \end{aligned}$
$\begin{aligned} t < \frac{6}{7} \end{aligned}$
$\begin{aligned} t > \frac{7}{6} \end{aligned}$

$\begin{aligned} \frac{12}{\sqrt[3]{- 27}} - 2 r > \frac{1}{3} - r \end{aligned}$

$\begin{aligned} r > \frac{13}{3} \end{aligned}$
$\begin{aligned} r < - \frac{13}{3} \end{aligned}$
$\begin{aligned} r < \frac{3}{13} \end{aligned}$
$\begin{aligned} r > - \frac{3}{13} \end{aligned}$

$\begin{aligned} \frac{1}{12} (m + 9) \geq \frac{1}{2} (\sqrt{6} - m) \end{aligned}$

$\begin{aligned} s = {m \in R, m \geq \frac{6 \sqrt{6} - 7}{9}} \end{aligned}$
$\begin{aligned} s = {m \in R, m \geq \frac{9 - 6 \sqrt{6}}{7}} \end{aligned}$
$\begin{aligned} s = {m \in R, m \geq \frac{7 - 6 \sqrt{6}}{9}} \end{aligned}$
$\begin{aligned} s = {m \in R, m \geq \frac{6 \sqrt{6} - 9}{7}} \end{aligned}$

مشاركة الدرس

فايسبوك

واتساب

تيليجرام

طباعة

السؤال

اختر الإجابة الصحيحة: حل المتباينات التالية في R باستعمال خواص المتباينات على الأعداد الحقيقية:

الحل

$\begin{aligned} \frac{- 7}{8} (\frac{8}{14} x + \frac{1}{\sqrt{7}}) \geq 0 \end{aligned}$

$\begin{aligned} x \leq \frac{4}{\sqrt{7}} \end{aligned}$
$\begin{aligned} x \leq \frac{\sqrt{7}}{4} \end{aligned}$
$\begin{aligned} x \geq \frac{4}{\sqrt{7}} \end{aligned}$
$\begin{aligned} x \leq \frac{- \sqrt{7}}{4} \end{aligned}$

[4-5] حل المتباينات الجبرية متعددة الخطوات في R

اختر الإجابة الصحيحة لكل مما يأتي:

حل المتباينات التالية في R باستعمال خواص المتباينات على الأعداد الحقيقية:

$5 (y + 1) \geq 8 - \sqrt[3]{- 125}$

$\begin{aligned} s = {y : y \in R, y \leq - \frac{5}{8}} \end{aligned}$
$\begin{aligned} s = {y : y \in R, y \leq \frac{8}{5}} \end{aligned}$
$\begin{aligned} s = {y : y \in R, y \geq \frac{5}{8}} \end{aligned}$
$\begin{aligned} s = {y : y \in R, y \geq \frac{8}{5}} \end{aligned}$

$\begin{aligned} \frac{1}{2} (x - \frac{4}{3}) + \frac{1}{6} x < - \frac{2}{3} \end{aligned}$

$x < \frac{4}{3}$
$x > 0$
$x < 0$
$x > \frac{- 4}{3}$

$\begin{aligned} \frac{z}{6} \leq 4 (\frac{1}{\sqrt{9}} - \frac{1}{3} z) \end{aligned}$

$z \leq - 4$
$z \geq - 4$
$z \geq 4$
$z \leq \frac{8}{9}$

$\begin{aligned} 2 (v - 4 \sqrt{5}) > 5 (v - \sqrt{5}) \end{aligned}$

$\begin{aligned} v < - \sqrt{5} \end{aligned}$
$\begin{aligned} v > \sqrt{5} \end{aligned}$
$\begin{aligned} v > - \sqrt{5} \end{aligned}$
$\begin{aligned} v < \sqrt{5} \end{aligned}$

$\begin{aligned} \frac{3}{2} n - \sqrt[3]{- 8} - \frac{1}{4} n \leq | - 5 | \end{aligned}$

$\begin{aligned} n < \frac{12}{5} \end{aligned}$
$\begin{aligned} n \leq \frac{12}{5} \end{aligned}$
$\begin{aligned} n \leq \frac{- 12}{5} \end{aligned}$
$\begin{aligned} n < \frac{- 12}{5} \end{aligned}$

$\begin{aligned} 12 h - \sqrt[3]{125} > 6 (h + 1) \end{aligned}$

$\begin{aligned} h > \frac{6}{11} \end{aligned}$
$\begin{aligned} h < \frac{11}{6} \end{aligned}$
$\begin{aligned} h > \frac{11}{6} \end{aligned}$
$\begin{aligned} h < \frac{6}{11} \end{aligned}$

$\begin{aligned} \frac{- 7}{8} (\frac{8}{14} x + \frac{1}{\sqrt{7}}) \geq 0 \end{aligned}$

$\begin{aligned} x \leq \frac{4}{\sqrt{7}} \end{aligned}$
$\begin{aligned} x \leq \frac{\sqrt{7}}{4} \end{aligned}$
$\begin{aligned} x \geq \frac{4}{\sqrt{7}} \end{aligned}$
$\begin{aligned} x \leq \frac{- \sqrt{7}}{4} \end{aligned}$

$\begin{aligned} 11 (2 - y) < 4 (5 + y) \end{aligned}$

$\begin{aligned} y > \frac{15}{2} \end{aligned}$
$\begin{aligned} y < \frac{2}{15} \end{aligned}$
$\begin{aligned} y > \frac{- 2}{15} \end{aligned}$
$\begin{aligned} y > \frac{2}{15} \end{aligned}$

$\begin{aligned} \frac{1}{5} (z - \sqrt{3}) \geq \frac{1}{5} (\sqrt{3} - z) \end{aligned}$

$\begin{aligned} z \geq \sqrt{3} \end{aligned}$
$\begin{aligned} z \leq \frac{2}{\sqrt{3}} \end{aligned}$
$\begin{aligned} z \leq \sqrt{3} \end{aligned}$
$\begin{aligned} z \geq \frac{2}{\sqrt{3}} \end{aligned}$

$\begin{aligned} \frac{t}{2} - \frac{1}{7} < \frac{t}{14} + \frac{5}{14} \end{aligned}$

$\begin{aligned} t > \frac{6}{7} \end{aligned}$
$\begin{aligned} t < \frac{7}{6} \end{aligned}$
$\begin{aligned} t < \frac{6}{7} \end{aligned}$
$\begin{aligned} t > \frac{7}{6} \end{aligned}$

$\begin{aligned} \frac{12}{\sqrt[3]{- 27}} - 2 r > \frac{1}{3} - r \end{aligned}$

$\begin{aligned} r > \frac{13}{3} \end{aligned}$
$\begin{aligned} r < - \frac{13}{3} \end{aligned}$
$\begin{aligned} r < \frac{3}{13} \end{aligned}$
$\begin{aligned} r > - \frac{3}{13} \end{aligned}$

$\begin{aligned} \frac{1}{12} (m + 9) \geq \frac{1}{2} (\sqrt{6} - m) \end{aligned}$

$\begin{aligned} s = {m \in R, m \geq \frac{6 \sqrt{6} - 7}{9}} \end{aligned}$
$\begin{aligned} s = {m \in R, m \geq \frac{9 - 6 \sqrt{6}}{7}} \end{aligned}$
$\begin{aligned} s = {m \in R, m \geq \frac{7 - 6 \sqrt{6}}{9}} \end{aligned}$
$\begin{aligned} s = {m \in R, m \geq \frac{6 \sqrt{6} - 9}{7}} \end{aligned}$

حلول الأسئلة

اختر الإجابة الصحيحة: حل المتباينات التالية في R باستعمال خواص المتباينات على الأعداد الحقيقية:

مشاركة الحل

[4-5] حل المتباينات الجبرية متعددة الخطوات في R

اختر الإجابة الصحيحة لكل مما يأتي:

حل المتباينات التالية في R باستعمال خواص المتباينات على الأعداد الحقيقية:

5(y+1)≥8−−1253

12(x−43)+16x<−23

z6≤4(19−13z)

2(v−45)>5(v−5)

32n−−83−14n≤|−5|

12h−1253>6(h+1)

−78(814x+17)≥0

11(2−y)<4(5+y)

15(z−3)≥15(3−z)

t2−17<t14+514

12−273−2r>13−r

112(m+9)≥12(6−m)

مشاركة الدرس

اختر الإجابة الصحيحة: حل المتباينات التالية في R باستعمال خواص المتباينات على الأعداد الحقيقية:

[4-5] حل المتباينات الجبرية متعددة الخطوات في R

اختر الإجابة الصحيحة لكل مما يأتي:

حل المتباينات التالية في R باستعمال خواص المتباينات على الأعداد الحقيقية:

5(y+1)≥8−−1253

12(x−43)+16x<−23

z6≤4(19−13z)

2(v−45)>5(v−5)

32n−−83−14n≤|−5|

12h−1253>6(h+1)

−78(814x+17)≥0

11(2−y)<4(5+y)

15(z−3)≥15(3−z)

t2−17<t14+514

12−273−2r>13−r

112(m+9)≥12(6−m)

$5 (y + 1) \geq 8 - \sqrt[3]{- 125}$

$\begin{aligned} \frac{1}{2} (x - \frac{4}{3}) + \frac{1}{6} x < - \frac{2}{3} \end{aligned}$

$\begin{aligned} \frac{z}{6} \leq 4 (\frac{1}{\sqrt{9}} - \frac{1}{3} z) \end{aligned}$

$\begin{aligned} 2 (v - 4 \sqrt{5}) > 5 (v - \sqrt{5}) \end{aligned}$

$\begin{aligned} \frac{3}{2} n - \sqrt[3]{- 8} - \frac{1}{4} n \leq | - 5 | \end{aligned}$

$\begin{aligned} 12 h - \sqrt[3]{125} > 6 (h + 1) \end{aligned}$

$\begin{aligned} \frac{- 7}{8} (\frac{8}{14} x + \frac{1}{\sqrt{7}}) \geq 0 \end{aligned}$

$\begin{aligned} 11 (2 - y) < 4 (5 + y) \end{aligned}$

$\begin{aligned} \frac{1}{5} (z - \sqrt{3}) \geq \frac{1}{5} (\sqrt{3} - z) \end{aligned}$

$\begin{aligned} \frac{t}{2} - \frac{1}{7} < \frac{t}{14} + \frac{5}{14} \end{aligned}$

$\begin{aligned} \frac{12}{\sqrt[3]{- 27}} - 2 r > \frac{1}{3} - r \end{aligned}$

$\begin{aligned} \frac{1}{12} (m + 9) \geq \frac{1}{2} (\sqrt{6} - m) \end{aligned}$

$5 (y + 1) \geq 8 - \sqrt[3]{- 125}$

$\begin{aligned} \frac{1}{2} (x - \frac{4}{3}) + \frac{1}{6} x < - \frac{2}{3} \end{aligned}$

$\begin{aligned} \frac{z}{6} \leq 4 (\frac{1}{\sqrt{9}} - \frac{1}{3} z) \end{aligned}$

$\begin{aligned} 2 (v - 4 \sqrt{5}) > 5 (v - \sqrt{5}) \end{aligned}$

$\begin{aligned} \frac{3}{2} n - \sqrt[3]{- 8} - \frac{1}{4} n \leq | - 5 | \end{aligned}$

$\begin{aligned} 12 h - \sqrt[3]{125} > 6 (h + 1) \end{aligned}$

$\begin{aligned} \frac{- 7}{8} (\frac{8}{14} x + \frac{1}{\sqrt{7}}) \geq 0 \end{aligned}$

$\begin{aligned} 11 (2 - y) < 4 (5 + y) \end{aligned}$

$\begin{aligned} \frac{1}{5} (z - \sqrt{3}) \geq \frac{1}{5} (\sqrt{3} - z) \end{aligned}$

$\begin{aligned} \frac{t}{2} - \frac{1}{7} < \frac{t}{14} + \frac{5}{14} \end{aligned}$

$\begin{aligned} \frac{12}{\sqrt[3]{- 27}} - 2 r > \frac{1}{3} - r \end{aligned}$

$\begin{aligned} \frac{1}{12} (m + 9) \geq \frac{1}{2} (\sqrt{6} - m) \end{aligned}$