جواب تمرین صفحه 31 درس 2 ریاضی نهم (عددهای حقیقی)

\(\begin{array}{l}\left| {a + b} \right| + 2\left| {a - b - c} \right| = \\\\\left| {0/25 - \frac{1}{4}} \right| + 2\left| {0/25 + \frac{1}{4} - 2\frac{1}{2}} \right| = \\\\\left| {\frac{1}{4} - \frac{1}{4}} \right| + 2\left| {\frac{1}{4} + \frac{1}{4} - 2\frac{1}{2}} \right| = \\\\0 + 2\left| {\frac{1}{2} - 2\frac{1}{2}} \right| = 2\left| { - 2} \right| = 2(2) = 4\end{array}\)

الف

\(\left| { - 3\sqrt 5 } \right| = - ( - 3\sqrt 5 ) = 3\sqrt 5 \)

ب

\(\begin{array}{l}\left| {7 - 5\sqrt 3 } \right| = \left| {\sqrt {49} - \sqrt {25 \times 3} } \right| = \\\\\left| {\sqrt {49} - \sqrt {75} } \right| = - (\sqrt {49} - \sqrt {75} ) = \\\\\sqrt {75} - \sqrt {49} = 5\sqrt 3 - 7\end{array}\)

ج

\(\left| {0 + \sqrt 5 } \right| = \left| {\sqrt 5 } \right| = \sqrt 5 \)

\(\begin{array}{l}\left| {5 - 12} \right| > 1 + \;\bigcirc \\\\ \Rightarrow \left| { - 7} \right| > 1 + \;\bigcirc \\\\ \Rightarrow 7 > 1 + \;\bigcirc \\\\ \Rightarrow 7 - 1 > 1 + \;\bigcirc - 1\\\\ \Rightarrow 6 > \;\bigcirc \end{array}\)

پس \(\bigcirc \) ، هر عدد کوچکتر از 6 می تواند باشد؛ مثل 5، 4، 2- و ...

\(\begin{array}{l}\left\{ \begin{array}{l}a = - 2\\\\\left| a \right| = \left| { - 2} \right| = 2\end{array} \right. \Rightarrow a + \left| a \right| = - 2 + 2 = 0\\\\\left\{ \begin{array}{l}a = 0\\\\\left| a \right| = \left| 0 \right| = 0\end{array} \right. \Rightarrow a + \left| a \right| = 0 + 0 = 0\\\\\left\{ \begin{array}{l}a = 2\\\\\left| a \right| = \left| 2 \right| = 2\end{array} \right. \Rightarrow a + \left| a \right| = 2 + 2 = 4\end{array}\)

خیر؛ به ازای هیچ یک از اعداد حقیقی، عبارت a+|a| منفی نخواهد شد.

\(\begin{array}{l}a = - 7\\\\\sqrt {{{( - 7)}^2}} = \sqrt {49} = 7 \ne - 7\end{array}\)

مثال فوق، یک مثال نقض برای تساوی \(\sqrt {{a^2}} = a\) می باشد.

\(\begin{array}{l}\sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} = \left| {\sqrt 2 - 1} \right| = \sqrt 2 - 1\\\\\sqrt {{{\left( {1 - \sqrt {10} } \right)}^2}} = \left| {1 - \sqrt {10} } \right| = - (1 - \sqrt {10} ) = \sqrt {10} - 1\end{array}\)