حل المعادلات التالية في R باستعمال الجذر التربيعي:

$n^2-5=20$

$n^2-5=20\Rightarrow n^2=25\Rightarrow n=5 or n=-5$

اختبار الفصل

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

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

$$\begin{gathered} z=\sqrt[3]{-8}+2z\Rightarrow z=-2+2z\Rightarrow 2=2z-z\Rightarrow z=2 \\ 2=\sqrt[3]{-8}+2\left(2\right)\Rightarrow 2=-2+4\Rightarrow 2=2 \end{gathered}$$

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

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

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

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

$3z÷21=\frac{1}{7}$

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

$\sqrt[3]{125}\times ÷|-9|=5^2$

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

$\sqrt{3}x÷9=\sqrt{3}÷3$

$$\begin{gathered} \sqrt{3}x÷9=\sqrt{3}÷3\Rightarrow \frac{\sqrt{3}x}{9}=\frac{\sqrt{3}}{3}\Rightarrow \frac{9}{\sqrt{3}}\left(\frac{\sqrt{3x}}{9}\right)=\frac{9}{\sqrt{3}}\times \frac{\sqrt{3}}{3} \\ \Rightarrow x=3 \end{gathered}$$

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

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

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

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

$6z÷13=5z÷13$

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

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

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

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

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

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

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

حل المعادلات التالية في R باستعمال الخواص:

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

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

$\sqrt[3]{-8}y÷|-8|=4\sqrt{2}$

$\begin{array}{rl}& \sqrt[3]{-8}y÷|-8|=4\sqrt{2}\Rightarrow \frac{-2y}{8}=4\sqrt{2}\Rightarrow -2y=32\sqrt{2}\\ & \Rightarrow y=\frac{32\sqrt{2}}{-2}\Rightarrow y=-16\sqrt{2}\end{array}$

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

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

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

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

$|y-12|=7$

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

مجموعة الحل هي: $\{19,5\}$

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

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

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: $\varnothing$

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

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

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

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

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

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

حل المعادلات التالية في R باستعمال الجذر التربيعي:

$x^2=64$

$x^2=64\Rightarrow x=8 or x=-8$

$9y^2=1$

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

$12t^2=4$

$$\begin{gathered} 12t^2=4\Rightarrow t^2=\frac{4}{12}\Rightarrow t=\frac{2}{2\sqrt{3}} or t=-\frac{2}{2\sqrt{3}} \\ \Rightarrow t=\frac{1}{\sqrt{3}} or t=-\frac{1}{\sqrt{3}} \end{gathered}$$

$n^2-5=20$

$n^2-5=20\Rightarrow n^2=25\Rightarrow n=5 or n=-5$

$7-z^2=-42$

$7-z^2=-42\Rightarrow -z^2=-42-7\Rightarrow z^2=49\Rightarrow z=7 or z=-7$

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

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

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

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

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

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

حل المعادلات التالية في R باستعمال خاصية الضرب الصفري:

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

$\begin{array}{rl}& (\mathrm{y}-4)(\mathrm{y}+4)=0\Rightarrow (\mathrm{y}-4)=0 or (\mathrm{y}+4)=0\\ & \Rightarrow \mathrm{y}=4 or \mathrm{y}=-4\end{array}$

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

$(z-7)(z-7)=0\Rightarrow z-7=0 or z-7=0\Rightarrow z=7$

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

$\begin{array}{rl}& (x+\sqrt{5})(x-\sqrt{3})=0\Rightarrow x+\sqrt{5}=0 or x-\sqrt{3}=0\\ & \Rightarrow x=-\sqrt{5} or x=\sqrt{3}\end{array}$

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

$\begin{array}{rl}& (\sqrt{2}-h)(\sqrt{2}+h)=0\Rightarrow \sqrt{2}-h=0 or \sqrt{2}+h=0\\ & \Rightarrow h=\sqrt{2} or h=-\sqrt{2}\end{array}$

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

$\begin{array}{rl}& (4\mathrm{t}+8)(3\mathrm{t}-7)=0\Rightarrow 4\mathrm{t}+8=0 or 3\mathrm{t}-7=0\\ & \mathrm{t}=-2 or \mathrm{t}=\frac{7}{3}\end{array}$

${\mathrm{z}}^2-\mathrm{z}=0$

$z^2-z=0\Rightarrow z(z-1)=0\Rightarrow z=0 or z=1$

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

$$\begin{gathered} \sqrt{8}{x}^2+2x=0\Rightarrow 2\sqrt{2}{x}^2+2x=0\Rightarrow 2x(\sqrt{2}x+1)=0 \\ \Rightarrow 2\mathrm{x}=0 or \sqrt{2}\mathrm{x}+1=0\Rightarrow \mathrm{x}=0 or \mathrm{x}=\frac{-1}{\sqrt{2}} \end{gathered}$$

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

$\begin{array}{rl}& 3\sqrt{7}{n}^2-3\sqrt{7}n=0\Rightarrow 3\sqrt{7}n(n-1)=0\Rightarrow 3\sqrt{7}n=0 or n-1=0\\ & \Rightarrow n=0 or n=1\end{array}$

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

$\begin{array}{rl}& 3\frac{1}{5}{y}^2-\frac{1}{5}y=0\Rightarrow \frac{1}{5}y(y-1)=0\Rightarrow \frac{1}{5}y=0 or y-1=0\\ & \Rightarrow y=0 or y=1\end{array}$

حل المتباينات التالية باستعمال خواص المتباينات على الأعداد الحقيقية:

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

$\begin{array}{rl}& 2(x-\sqrt{2})\ge x-\sqrt{2}\Rightarrow 2x-\sqrt{2}\ge x-\sqrt{2}\\ & \Rightarrow 2x-x\ge 2\sqrt{2}-\sqrt{2}\Rightarrow \ge x\ge \sqrt{2}\end{array}$

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

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

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

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

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

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

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

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

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

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

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

$$\begin{gathered} \frac{1}{6}(z-\frac{12}{5})+\frac{5}{6}z\ge \frac{-3}{5}\Rightarrow \frac{1}{6}z+\frac{5}{6}z-\frac{2}{5}\ge \frac{-3}{5} \\ \Rightarrow \frac{1+5}{6}z\ge \frac{-3}{5}+\frac{2}{5}\Rightarrow \frac{6}{6}z\ge \frac{-3+2}{5}\Rightarrow z\ge \frac{-1}{5} \end{gathered}$$

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

$$\begin{gathered} \frac{y}{4}\le 2(\frac{1}{\sqrt{16}}-\frac{1}{8}y)\Rightarrow \frac{y}{4}\le \frac{2}{4}-\frac{1}{4}y\Rightarrow \frac{y}{4}+\frac{y}{4}\le \frac{1}{2} \\ \Rightarrow \frac{1}{2}y\le \frac{1}{2}\Rightarrow y\le 1 \end{gathered}$$

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

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

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

$$\begin{gathered} \frac{1}{3}(h-\sqrt{2})\le \frac{2}{3}(\sqrt{2}-h)\Rightarrow \frac{1}{3}h-\frac{\sqrt{2}}{3}\le \frac{2\sqrt{2}}{3}-\frac{2}{3}h \\ \Rightarrow \frac{1}{3}h+\frac{2}{3}h\le \frac{2\sqrt{2}}{3}+\frac{\sqrt{2}}{3}\Rightarrow \frac{3}{3}h\le \frac{3\sqrt{2}}{3}\Rightarrow h\le \sqrt{2} \end{gathered}$$

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

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

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

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