6.1: إيجاد قيمة التعبيرات
- Page ID
- 167093
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)القيمة المطلقة للعدد الحقيقي\(a\)، المكتوبة\(|a|\)، هي المسافة من\(a\) إلى\(0\) على خط الأعداد.
للبحث\(|−4|\)، اسأل: «ما هي المسافة من\(−4\) إلى\(0\)؟». ارسم خط أرقام وشاهد ذلك\(|−4| = 4\). وبالمثل\(|4| = 4\)، كما هو موضح في الشكل أدناه.
قم بتقييم التعبيرات التالية:
- \(|8−2|− |4−7|\)
- \(5|−3|+|−9|^2\)
- \(\dfrac{3}{5}|6 + (−3)^3|\)
- \(\left|\dfrac{(−2)^2 + 12}{3} +5 \right|+|−4+2|\)
الحل
- للتقييم\(|8 − 2| − |4 − 7|\)، قم أولاً بالتبسيط داخل القيمة المطلقة.
\(\begin{array} &&|8 − 2| − |4 − 7| &\text{Given} \\ &= |6| − |− 3| &\text{Simplify inside the absolute value} \\ &= (6) − (3) &\text{Absolute value definition} \\ &= 3 & \end{array}\)
- أولاً، قم بتبسيط القيم المطلقة، ثم قم بتطبيق العملية الحسابية المطلوبة.
\(\begin{array} &&5| − 3| + | − 9|^2 &\text{Given} \\ &= 5(3) + (9)^2 &\text{Absolute value definition} \\ &= 15 + 81 &\text{Simplify} \\ &= 96 & \end{array}\)
- استخدم ترتيب العمليات «PEMDAS» للتبسيط داخل القيمة المطلقة.
\(\begin{array} &&\dfrac{3}{5}|6 + (−3)^3| &\text{Given} \\ &=\dfrac{3}{5}|6 + (−27)| &\text{Evaluate the exponent term} \\ &= \dfrac{3}{5} − 21 &\text{Simplify inside the absolute value} \\ &= \dfrac{3}{5} (21) &\text{Absolute value definition} \\ &= \dfrac{63}{5} & \end{array}\)
- لتقييم التعبير في هذا الجزء، قم أولاً بتطبيق ترتيب التشغيل «PEMDAS» داخل القيمة المطلقة للتبسيط.
\(\begin{array} &&\left|\dfrac{(−2)^2 + 12}{3} +5 \right|+|−4+2| &\text{Given} \\ &= \left|\dfrac{(4 + 12)}{3} +5 \right|+|−2| &\text{Simplify} \\ &= \left|\dfrac{16}{3} +5 \right|+|−2| &\text{Note that \(3\)هي شاشة LCD الخاصة بـ\(\dfrac{16}{3}\) و\(5\). \(5\)يمكن كتابتها كـ\(\dfrac{5}{1}\)}\\ &=\ يسارة|\ dfrac {16} {3} +\ dfrac {5 (3)} {1 (3)}\\ اليمين |+|−2|\ النص {اضرب البسط والمقام على شاشة LCD لإضافة المصطلحات داخل القيمة المطلقة.}\\\ &=\ يسار|\ dfrac {31} {3}\ اليمين |+−2|\\ &=\ يسار (\ dfrac {31} {3}\ يمين) + (2) &\(\dfrac{5}{1}\) amp;\ text {تعريف القيمة المطلقة}\\ &=\ dfrac {31} {3} + 2 &\ text {على غرار ما ورد أعلاه،\(3\) هي شاشة LCD الخاصة بـ\(\dfrac{31}{3}\) و\(2\). \(2\)يمكن كتابتها كـ\(\dfrac{2}{1}\).}\\ &=\ dfrac {31} {3} +\ dfrac {2 (3)} {1 (3)}\\ text {اضرب\(\dfrac{2}{1}\)\(\dfrac{3}{3}\) لإضافة المصطلحين.}\\ &=\ dfrac {37} {3} و\ end {مصفوفة}\)
قم بتقييم التعبيرات المعطاة:
- \(|8 − 15|\)
- \(|− 3 −12|\)
- \(\left|− 2 + 11 − \left( −\dfrac{6}{4} \right) \right|\)
- \(\left|−\dfrac{1 + 5}{12} − 5\right|− 1\)
- \(|2 (5 + 6) − 20|\)
- \(\left|\dfrac{1}{2} (21 − 5) − |(−2)^3 \right|\)
- \(\left|−5 |− 2(−13 + 10) \right|\)
- \(\dfrac{3}{2} \left| 12 \left( \dfrac{−7 + 17}{(6 − 2)} \right) \right| + |− (−2)|\)