[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:

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

• $\begin{array}{rl}& s=\{y:y\in R,y\le -\frac{5}{8}\}\end{array}$
• $\begin{array}{rl}& \mathrm{s}=\{\mathrm{y}:\mathrm{y}\in \mathrm{R},\mathrm{y}\le \frac{8}{5}\}\end{array}$
• $\begin{array}{rl}& s=\{y:y\in R,y\ge \frac{5}{8}\}\end{array}$
• $\begin{array}{rl}& \mathbf{s}\mathbf{=}\mathbf{\{}\mathbf{y}\mathbf{:}\mathbf{y}\mathbf{\in }\mathbf{R}\mathbf{,}\mathbf{y}\mathbf{\ge }\frac{\mathbf{8}}{\mathbf{5}}\mathbf{\}}\end{array}$

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

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

$\begin{array}{rl}& \frac{z}{6}\le 4(\frac{1}{\sqrt{9}}-\frac{1}{3}z)\end{array}$

• $z\le -4$
• $\mathrm{z}\ge -4$
• $\mathrm{z}\ge 4$
• $\mathit{z}\mathbf{\le }\frac{\mathbf{8}}{\mathbf{9}}$

$\begin{array}{rl}& 2(v-4\sqrt{5})>5(v-\sqrt{5})\end{array}$

• $\begin{array}{rl}& \mathbf{v}\mathbf{<}\mathbf{-}\sqrt{\mathbf{5}}\end{array}$
• $\begin{array}{rl}& v>\sqrt{5}\end{array}$
• $\begin{array}{rl}& v>-\sqrt{5}\end{array}$
• $\begin{array}{rl}& v<\sqrt{5}\end{array}$

$\begin{array}{rl}& \frac{3}{2}n-\sqrt[3]{-8}-\frac{1}{4}n\le |-5|\end{array}$

• $\begin{array}{rl}& n<\frac{12}{5}\end{array}$
• $\begin{array}{rl}& \mathbf{n}\mathbf{\le }\frac{\mathbf{12}}{\mathbf{5}}\end{array}$
• $\begin{array}{rl}& n\le \frac{-12}{5}\end{array}$
• $\begin{array}{rl}& n<\frac{-12}{5}\end{array}$

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

• $\begin{array}{rl}& h>\frac{6}{11}\end{array}$
• $\begin{array}{rl}& h<\frac{11}{6}\end{array}$
• $\begin{array}{rl}& \mathbf{h}\mathbf{>}\frac{\mathbf{11}}{\mathbf{6}}\end{array}$
• $\begin{array}{rl}& h<\frac{6}{11}\end{array}$

$\begin{array}{rl}& \frac{-7}{8}(\frac{8}{14}x+\frac{1}{\sqrt{7}})\ge 0\end{array}$

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

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

• $\begin{array}{rl}& y>\frac{15}{2}\end{array}$
• $\begin{array}{rl}& y<\frac{2}{15}\end{array}$
• $\begin{array}{rl}& y>\frac{-2}{15}\end{array}$
• $\begin{array}{rl}& \mathbf{y}\mathbf{>}\frac{\mathbf{2}}{\mathbf{15}}\end{array}$

$\begin{array}{rl}& \frac{1}{5}(z-\sqrt{3})\ge \frac{1}{5}(\sqrt{3}-z)\end{array}$

• $\begin{array}{rl}& \mathbf{z}\mathbf{\ge }\sqrt{\mathbf{3}}\end{array}$
• $\begin{array}{rl}& z\le \frac{2}{\sqrt{3}}\end{array}$
• $\begin{array}{rl}& z\le \sqrt{3}\end{array}$
• $\begin{array}{rl}& z\ge \frac{2}{\sqrt{3}}\end{array}$

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

• $\begin{array}{rl}& t>\frac{6}{7}\end{array}$
• $\begin{array}{rl}& \mathbf{t}\mathbf{<}\frac{\mathbf{7}}{\mathbf{6}}\end{array}$
• $\begin{array}{rl}& t<\frac{6}{7}\end{array}$
• $\begin{array}{rl}& t>\frac{7}{6}\end{array}$

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

• $\begin{array}{rl}& r>\frac{13}{3}\end{array}$
• $\begin{array}{rl}& \mathbf{r}\mathbf{<}\mathbf{-}\frac{\mathbf{13}}{\mathbf{3}}\end{array}$
• $\begin{array}{rl}& r<\frac{3}{13}\end{array}$
• $\begin{array}{rl}& r>-\frac{3}{13}\end{array}$

$\begin{array}{rl}& \frac{1}{12}(\mathrm{m}+9)\ge \frac{1}{2}(\sqrt{6}-\mathrm{m})\end{array}$

• $\begin{array}{rl}& \mathrm{s}=\{\mathrm{m}\in \mathrm{R},\mathrm{m}\ge \frac{6\sqrt{6}-7}{9}\}\end{array}$
• $\begin{array}{rl}& \mathrm{s}=\{\mathrm{m}\in \mathrm{R},\mathrm{m}\ge \frac{9-6\sqrt{6}}{7}\}\end{array}$
• $\begin{array}{rl}& \mathrm{s}=\{\mathrm{m}\in \mathrm{R},\mathrm{m}\ge \frac{7-6\sqrt{6}}{9}\}\end{array}$
• $\begin{array}{rl}& \mathbf{s}\mathbf{=}\mathbf{\{}\mathbf{m}\mathbf{\in }\mathbf{R}\mathbf{,}\mathbf{m}\mathbf{\ge }\frac{\mathbf{6}\sqrt{\mathbf{6}}\mathbf{-}\mathbf{9}}{\mathbf{7}}\mathbf{\}}\end{array}$