نمونه سوالات تشریحی فصل 2 ریاضی نهم همراه با جواب تشریحی

نمایش به صورت مجموعه با اعضا:

\([ + 7,\infty )\)

نمایش به صورت نماد های ریاضی:

\(\{ x|x \in \mathbb{R},x \ge 7\} \)

نمایش روی محور:

\(\begin{array}{l}1) - \frac{2}{7} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{3}{5} \Rightarrow [7,5] = 35\\\\ \Rightarrow \frac{{ - 10}}{{35}} < \frac{1}{{35}} < \frac{2}{{35}} < \frac{3}{{35}} < \frac{4}{{35}} < \frac{{21}}{{35}} = \\\\ \Rightarrow \frac{{ - 2}}{7} < \frac{1}{{35}} < \frac{2}{{35}} < \frac{3}{{35}} < \frac{4}{{35}} < \frac{3}{5}\end{array}\)

\(\begin{array}{l}2)\frac{1}{8} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{{}}{{}} < \frac{1}{3} \Rightarrow [8,3] = 24\\\\ \Rightarrow \frac{3}{{24}} < \frac{4}{{24}} < \frac{5}{{24}} < \frac{6}{{24}} < \frac{7}{{24}} < \frac{8}{{24}} = \\\\ \Rightarrow \frac{1}{8} < \frac{1}{6} < \frac{5}{{24}} < \frac{1}{4} < \frac{7}{{24}} < \frac{1}{3}\end{array}\)

اعداد گویا عبارتند از:

\(\frac{{\sqrt {343} }}{{\sqrt 7 }},0,\frac{{1 + 3 \times 4}}{{8 - 2}},\frac{{{4^2} - 7 \times 8 + 1}}{{\sqrt {25} - \sqrt {16} }}\)

\(\frac{{\sqrt 4 }}{{\sqrt {25} }} < \frac{5}{9} < \frac{5}{8} < \frac{3}{4} < \frac{8}{5}\)

\(\begin{array}{l}1)3/4678678678... = 3/4\overline {678} \\\\2)6/023023023... = 6/\overline {023} \\\\3)81/111... = 81/\overline {01} \end{array}\)

\(\begin{array}{l}1)1/\overline {43} = 1/434343...\,\,,\,\,1/4\overline 3 = 1/4333...\\\\ \Rightarrow 1/\overline {43} > 1/4\overline 3 \\\\2)8/0\overline {101} = 8/0101101101...\,\,,\,\,8/\overline {01} = 8/010101...\\\\ \Rightarrow 8/0\overline {101} > 8/\overline {01} \end{array}\)

مختوم\(1)\frac{6}{8} = \frac{3}{4} = 0/75 \Rightarrow \)

متناوب ساده\(2)\frac{{19}}{{57}} = \frac{1}{3} = 0/\overline 3 \Rightarrow \)

متناوب مرکب\(3)\frac{{130}}{{1200}} = \frac{{13}}{{120}} = \frac{{13}}{{12 \times 2 \times 5}} \Rightarrow \)

متناوب ساده\(4)\frac{{24}}{{26}} = \frac{{12}}{{13}} \Rightarrow \)

متناوب مرکب\(5)\frac{{102}}{{36}} = \frac{{51}}{{18}} = \frac{{17}}{6} \Rightarrow \)

\(\begin{array}{l}1)7/1\overline 4 + 1/\overline 3 = 7/1\overline 4 + 1/3\overline 3 = 7/4\overline 7 \\\\2)8/\overline 8 - 3/\overline {35} = 8/\overline {88} - 3/\overline {35} = 5/\overline {53} \end{array}\)

\(\frac{1}{{1024}},\frac{5}{2},\frac{1}{{40}}\)

\(\frac{2}{{11}},\frac{1}{3},\frac{1}{7}\)

\(\frac{2}{{11}} = 0/181818 = 0/\overline {18} \)

رقم هفتم برابر 1 و رقم بیستم برابر 8 می باشد.

\(\begin{array}{l}2\frac{1}{3} - 3\frac{1}{2} \div (4\frac{1}{5} \div \frac{3}{4}) = \\\\ \Rightarrow \frac{7}{3} - \frac{7}{2} \div (\frac{{21}}{5} \div \frac{3}{4}) = \frac{7}{3} - \frac{7}{2} \div (\frac{{21}}{5} \times \frac{4}{3}) = \\\\ \Rightarrow \frac{7}{3} - \frac{7}{2} \div (\frac{{28}}{5}) = \frac{7}{3} - \frac{7}{2} \times \frac{5}{{28}} = \frac{7}{3} - \frac{5}{8} = \\\\ \Rightarrow \frac{{56 - 15}}{{24}} = \frac{{41}}{{24}} = 1\frac{{17}}{{24}}\end{array}\)

\(\begin{array}{l}\frac{{( - \frac{{24}}{{10}} + \frac{5}{4}) + (\frac{9}{8} + \frac{{ - 12}}{5})}}{{2(\frac{3}{5} + \frac{9}{{40}}) - 3(\frac{1}{5} + \frac{2}{3})}} = \\\\ \Rightarrow \frac{{(\frac{{ - 96 + 50}}{{40}}) + (\frac{{45 - 96}}{{40}})}}{{2(\frac{{24 + 9}}{{40}}) - 3(\frac{{3 + 10}}{{15}})}} = \\\\ \Rightarrow \frac{{\frac{{ - 96 + 50 + 45 - 96}}{{40}}}}{{\frac{{24 + 9}}{{20}} - \frac{{13}}{5}}} = \\\\ \Rightarrow \frac{{ - \frac{{97}}{{40}}}}{{\frac{{33 - 52}}{{20}}}} = \frac{{\frac{{ - 97}}{{40}}}}{{ - \frac{{19}}{{20}}}} = \frac{{ - 97 \times 20}}{{ - 19 \times 40}} = \frac{{97}}{{38}}\end{array}\)

\(\begin{array}{l}\frac{{( - \frac{{24}}{{10}} + \frac{5}{4}) + (\frac{9}{8} + \frac{{ - 12}}{5})}}{{2(\frac{3}{5} + \frac{9}{{40}}) - 3(\frac{1}{5} + \frac{2}{3})}} = \\\\ \Rightarrow \frac{{(\frac{{ - 96 + 50}}{{40}}) + (\frac{{45 - 96}}{{40}})}}{{2(\frac{{24 + 9}}{{40}}) - 3(\frac{{3 + 10}}{{15}})}} = \\\\ \Rightarrow \frac{{\frac{{ - 96 + 50 + 45 - 96}}{{40}}}}{{\frac{{24 + 9}}{{20}} - \frac{{13}}{5}}} = \\\\ \Rightarrow \frac{{ - \frac{{97}}{{40}}}}{{\frac{{33 - 52}}{{20}}}} = \frac{{\frac{{ - 97}}{{40}}}}{{ - \frac{{19}}{{20}}}} = \frac{{ - 97 \times 20}}{{ - 19 \times 40}} = \frac{{97}}{{38}}\end{array}\)

\(\begin{array}{l}2 + \frac{{2 + \frac{{2 + \frac{1}{2}}}{2}}}{{2 - \frac{{2 - \frac{1}{2}}}{2}}} = 2 + \frac{{2 + \frac{{\frac{5}{2}}}{2}}}{{2 - \frac{{\frac{3}{2}}}{2}}} = 2 + \frac{{2 + \frac{5}{4}}}{{2 - \frac{3}{4}}} = \\\\ \Rightarrow 2 + \frac{{\frac{{13}}{4}}}{{\frac{5}{4}}} = 2 + \frac{{13}}{5} = \frac{{10 + 13}}{5} = \frac{{23}}{5}\end{array}\)

\(\begin{array}{l}\frac{y}{x} = \frac{{80}}{{48}} = \frac{5}{3} \Rightarrow 3y = 5x\\\\y = x + 250 \Rightarrow \times 3 \Rightarrow 3y = 3x + 750\\\\ \Rightarrow 5x = 3x + 750 \Rightarrow 5x - 3x = 750\\\\ \Rightarrow 2x = 750 \Rightarrow x = \frac{{750}}{2} = 375\\\\ \Rightarrow y = x + 250 \Rightarrow y = 375 + 250 = \\\\ \Rightarrow y = 625\end{array}\)

الف)

\(4 < \sqrt {17} < 5 \Rightarrow - 5 < - \sqrt {17} < - 4\)

ب)

\(\begin{array}{l}4 < \sqrt {18} < 5 \Rightarrow 5 + 4 < 5 + \sqrt {18} < 5 + 5\\\\ \Rightarrow 9 < 5 + \sqrt {18} < 10\end{array}\)

)پ

\(\begin{array}{l}5 < \sqrt {28} < 6 \Rightarrow - 6 < - \sqrt {28} < - 5\\\\ \Rightarrow 7 - 6 < 7 - \sqrt {28} < 7 - 5\\\\ \Rightarrow 1 < 7 - \sqrt {28} < 2\end{array}\)

ت)

\(\begin{array}{l}2\sqrt 5 = \sqrt {20} \Rightarrow 4 < \sqrt {20} < 5\\\\ \Rightarrow 4 < 2\sqrt 5 < 5 \Rightarrow - 5 < - 2\sqrt 5 < - 4\\\\ \Rightarrow 3 - 5 < 3 - 2\sqrt 5 < 3 - 4\\\\ \Rightarrow - 2 < 3 - 2\sqrt 5 < - 1\end{array}\)

\(A = - 2 - \sqrt 6 \)

\(\begin{array}{l}1)|5\sqrt 5 - 7\sqrt 5 | - |7\sqrt 6 - 6\sqrt 7 | = \\\\ \Rightarrow - (5\sqrt 7 - 7\sqrt 5 ) - (7\sqrt 6 - 6\sqrt 7 ) = \\\\ \Rightarrow - 5\sqrt 7 + 7\sqrt 5 - 7\sqrt 6 + 6\sqrt 7 = \\\\ \Rightarrow \sqrt 7 + 7(\sqrt 5 - \sqrt 6 ) = \sqrt 7 - 7(\sqrt 6 - \sqrt 5 )\\\\2)|10 - 4\sqrt 5 | - |4\sqrt 5 - 9| = \\\\ \Rightarrow (10 - 4\sqrt 5 ) + (9 - 4\sqrt 5 ) = 19 - 8\sqrt 5 \end{array}\)

\(\begin{array}{l}1)a < 0,\frac{{ - \sqrt {{a^2}} }}{{ - |a|}} = \frac{{ - |a|}}{{ - |a|}} = \frac{{|a|}}{{|a|}} = 1\\\\2)a > 0,\frac{{|a|}}{{ - a}} = \frac{a}{{ - a}} = - 1\end{array}\)

\(\begin{array}{l}x < - 3 \Rightarrow x + 3 < 0\\\\1)\sqrt {{{(x + 3)}^2}} + \sqrt {{{( - 3 - x)}^2}} = \\\\ \Rightarrow |x + 3| + | - 3 - x| = |x + 3| + |x + 3| = \\\\ \Rightarrow 2|x + 3| = - 2(x + 3) = - 2x - 6\\\\2)\sqrt {{{(|x + 3| + | - 3 - x|)}^2}} = \\\\ \Rightarrow \sqrt {{{(|x + 3| + |x + 3|)}^2}} = \sqrt {{{(2|x + 3|)}^2}} = \\\\ \Rightarrow |2|x + 3|| = 2|x + 3| = \\\\ \Rightarrow - 2(x + 3) = - 2x - 6\end{array}\)

\(\frac{{11}}{{112}}\)

\(2/\overline 3 \)

2 و 5

گنگ

مجموعه اعداد گویا

\(\frac{{136}}{{99}}\)

گنگ

1 (یک)

5-

گنگ

6- و 5-

گویا

6

گویا

مثبت

\( - 7 - \sqrt {11} + 3\sqrt 2 \)

منفی

b-a

مبدأ

اجتماع