[4-4] حل المتباينات الجبرية ذات الخطوتين في R

اختر الإجابة الصحيحة لكل مما يأتي:

حل المتباينات التالية في R باستعمال خواص الجمع والطرح:

$4(x-\sqrt{3})<3x+\sqrt{3}$

• $\begin{array}{rl}& \mathbf{s}\mathbf{=}\mathbf{\{}\mathbf{x}\mathbf{\in }\mathbf{R}\mathbf{,}\mathbf{x}\mathbf{<}\mathbf{5}\sqrt{\mathbf{3}}\mathbf{\}}\end{array}$
• $\begin{array}{rl}& \mathrm{s}=\{\mathrm{x}\in \mathrm{R},\mathrm{x}>5\sqrt{3}\}\end{array}$
• $\begin{array}{rl}& s=\{x\in R,x\le 5\sqrt{3}\}\end{array}$
• $\begin{array}{rl}& \mathrm{s}=\{\mathrm{x}\in \mathrm{R},\mathrm{x}\ge 5\sqrt{3}\}\end{array}$

$\begin{array}{rl}& 3y-\sqrt[3]{8}\ge 4y+\sqrt[3]{-27}\end{array}$

• $\begin{array}{rl}& y\le -1\end{array}$
• $\begin{array}{rl}& y\ge -1\end{array}$
• $y\ge 1$
• $\begin{array}{rl}& \mathbf{y}\mathbf{\le }\mathbf{1}\end{array}$

$\begin{array}{rl}& 4(\frac{1}{4}z+\frac{5}{14})<0\end{array}$

• $\begin{array}{rl}& \mathbf{z}\mathbf{<}\mathbf{-}\frac{\mathbf{10}}{\mathbf{7}}\end{array}$
• $\begin{array}{rl}& z<\frac{10}{7}\end{array}$
• $z>\frac{10}{7}$
• $\begin{array}{rl}& z<-\frac{10}{7}\end{array}$

$\begin{array}{rl}& 11(v-1)>10(v-2)\end{array}$

• $\begin{array}{rl}& s=\{v\in R,v<-9\}\end{array}$
• $\begin{array}{rl}& \mathbf{s}\mathbf{=}\mathbf{\{}\mathbf{v}\mathbf{\in }\mathbf{R}\mathbf{,}\mathbf{v}\mathbf{>}\mathbf{-}\mathbf{9}\mathbf{\}}\end{array}$
• $\begin{array}{rl}& s=\{v\in R,v\le -9\}\end{array}$
• $\begin{array}{rl}& \mathrm{s}=\{\mathrm{v}\in \mathrm{R},\mathrm{v}\ge -9\}\end{array}$

حل المتباينات التالية في R باستعمال خواص الضرب والقسمة:

$\frac{4h}{6}\ge \frac{-8}{21}$

• $\begin{array}{rl}& \mathrm{s}=\{\mathrm{h}:\mathrm{h}\in \mathrm{R},\mathrm{h}\le \frac{4}{7}\}\end{array}$
• $\begin{array}{rl}& \mathrm{s}=\{\mathrm{h}:\mathrm{h}\in \mathrm{R},\mathrm{h}\le -\frac{4}{7}\}\end{array}$
• $\begin{array}{rl}& \mathbf{s}\mathbf{=}\mathbf{\{}\mathbf{h}\mathbf{:}\mathbf{h}\mathbf{\in }\mathbf{R}\mathbf{,}\mathbf{h}\mathbf{\ge }\mathbf{-}\frac{\mathbf{4}}{\mathbf{7}}\mathbf{\}}\end{array}$
• $\begin{array}{rl}& \mathrm{s}=\{\mathrm{h}:\mathrm{h}\in \mathrm{R},\mathrm{h}\ge \frac{4}{7}\}\end{array}$

$\begin{array}{rl}& \frac{\sqrt{3}x}{5}<\frac{-6}{7}\end{array}$

• $x>\frac{10\sqrt{3}}{7}$
• $x\le \frac{-10\sqrt{3}}{7}$
• $x\ge \frac{-10\sqrt{3}}{7}$
• $\mathit{x}\mathbf{<}\frac{\mathbf{-}\mathbf{10}\sqrt{\mathbf{3}}}{\mathbf{7}}$

$\begin{array}{rl}& \frac{1}{\sqrt{2}}\le \frac{\sqrt{2}n}{9}\end{array}$

• $\mathrm{n}\le \frac{9}{2}$
• $\mathit{n}\mathbf{\ge }\frac{\mathbf{9}}{\mathbf{2}}$
• $n\ge \frac{-9}{2}$
• $n\le \frac{-9}{2}$

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

$\begin{array}{rl}& 5y+\sqrt[3]{-27}>3y-\sqrt[3]{8}\end{array}$

• $y\le \frac{1}{2}$
• $y\le \frac{-1}{2}$
• $y\ge \frac{-1}{2}$
• $\mathit{y}\mathbf{>}\frac{\mathbf{1}}{\mathbf{2}}$

$\begin{array}{rl}& 4(\frac{1}{7}-\frac{3}{12}z)\le 0\end{array}$

• $\begin{array}{rl}& \mathrm{z}\ge \frac{-4}{7}\end{array}$
• $\begin{array}{rl}& \mathrm{z}\le \frac{7}{4}\end{array}$
• $\begin{array}{rl}& \mathbf{z}\mathbf{\le }\frac{\mathbf{4}}{\mathbf{7}}\end{array}$
• $\begin{array}{rl}& z\ge \frac{4}{7}\end{array}$

$\begin{array}{rl}& \frac{3k}{2}-\frac{1}{3}\ge \frac{2}{3}-1\end{array}$

• $\begin{array}{rl}& \mathrm{k}\ge \frac{2}{3}\end{array}$
• $\begin{array}{rl}& \mathbf{k}\mathbf{\ge }\mathbf{0}\end{array}$
• $\begin{array}{rl}& \mathrm{k}\le \frac{2}{3}\end{array}$
• $\begin{array}{rl}& k\le 0\end{array}$

$\begin{array}{rl}& \frac{1}{5}(11x+20)<2\end{array}$

• $x>\frac{-10}{11}$
• $x>\frac{10}{11}$
• $\mathit{x}\mathbf{<}\frac{\mathbf{-}\mathbf{10}}{\mathbf{11}}$
• $x<\frac{10}{11}$