Class 9 Maths Chapter 1 সংখ্যা প্ৰণালী গণিতৰ অভ্যাস পুস্তিকাৰ সমাধান
গণিতৰ অভ্যাস পুস্তিকাৰ সমাধান ঃ নৱম শ্ৰেণী
প্ৰশ্ন নং-1 ৰ পৰা প্ৰশ্ন নং-12 লৈ সমাধান পাবৰ বাবে ইয়াত ক্লিক কৰা- Click Here.
13. (i) 4 আৰু 5 ৰ মাজত থকা 5 টা পৰিমেয় সংখ্যা উলিওৱা।
Solution:
\(\LARGE 4=\frac{4}{1}=\frac{4×6}{1×6}=\frac{24}{6}\)
\(\LARGE 5=\frac{5}{1}=\frac{5×6}{1×6}=\frac{30}{6}\)
[Hint: যিহেতু মাজত থকা 5 টা পৰিমেয় সংখ্যা উলিয়াব দিছে, গতিকে 5+1=6 ৰে দুয়োটা ভগ্নাংশৰে লৱ আৰু হৰক পূৰণ কৰা হৈছে।]
গতিকে, 4 আৰু 5 ৰ মাজত থকা 5 টা পৰিমেয় সংখ্যা হ'ল- \(\LARGE \frac{25}{6}, \frac{26}{6}, \frac{27}{6}, \frac{28}{6}, \frac{29}{6}\)
13. (ii) \(\LARGE \frac{1}{3}\) আৰু \( \LARGE \frac{2}{3}\) ৰ মাজত থকা 5 টা পৰিমেয় সংখ্যা উলিওৱা।
Solution:
\(\LARGE \frac{1}{3}=\frac{1×6}{3×6}=\frac{6}{18}\)
\(\LARGE \frac{2}{3}=\frac{2×6}{3×6}=\frac{12}{18}\)
[Hint: যিহেতু মাজত থকা 5 টা পৰিমেয় সংখ্যা উলিয়াব দিছে, গতিকে 5+1=6 ৰে দুয়োটা ভগ্নাংশৰে লৱ আৰু হৰক পূৰণ কৰা হৈছে।]
গতিকে, \(\LARGE \frac{1}{3}\) আৰু \( \LARGE \frac{2}{3}\) ৰ মাজত থকা 5 টা পৰিমেয় সংখ্যা হ'ল- \(\LARGE \frac{7}{18}, \frac{8}{18}, \frac{9}{18}, \frac{10}{18}, \frac{11}{18}\)
14. \(\LARGE \frac{p}{q}\) আৰ্হিত প্ৰকাশ কৰা ( p, q অখণ্ড আৰু q≠0)
a) \(0.\overline{7}\)
b) \(0.\overline{38}\)
c) \(0.\overline{231}\)
Solution:
a) \(0.\overline{7}\) ক \(\LARGE \frac{p}{q}\) আৰ্হিত প্ৰকাশ
ধৰাহল,
x=0.7777.............(i)
10x=7.7777.......(ii) [দুয়োপক্ষত 10 পূৰণ কৰিলে]
এতিয়া, (ii)-(i),
10x - x = (7.7777.......) - (0.7777.......)
⇒ 9x = 7
⇒x = \(\LARGE \frac{7}{9}\)
গতিকে, \(0.\overline{7}\)=\(\LARGE \frac{7}{9}\)
b) \(0.\overline{38}\) ক \(\LARGE \frac{p}{q}\) আৰ্হিত প্ৰকাশ
ধৰাহল,
x = 0.383838...... (i)
100x = 38.383838...... (ii) [দুয়ো পক্ষত 100 পূৰণ কৰিলে]
এতিয়া, (ii) - (i),
100x-x = (38.383838.......) - (0.383838......)
⇒99x = 38
⇒x= \(\large \frac{38}{99}\)
গতিকে, \(0.\overline{38}\)=\(\large \frac{38}{99}\)
c) \(0.\overline{231}\) ক \(\LARGE \frac{p}{q}\) আৰ্হিত প্ৰকাশ
ধৰাহল,
x = 0.231231...... (i)
1000x = 231.231231...... (ii) [দুয়ো পক্ষক 1000 ৰে পূৰণ কৰিলে]
এতিয়া, (ii) - (i),
1000x-x = (231.231231.......) - (0.231231......)
⇒999x = 231
⇒x= \(\large \frac{231}{999}\)
গতিকে, \(0.\overline{231}\)=\(\large\frac{231}{999}\)
15. সৰল কৰা।
(i) \(\sqrt{2}+2\sqrt{2}+3\sqrt{2}\)
(ii) \(5\sqrt{3}-2\sqrt{3}+\sqrt{3}\)
(iii) \((\sqrt{3}+1)(\sqrt{3}-1)\)
(iv) \((\sqrt{7}+3)^2\)
(v) \((2-\sqrt{5})^2\)
(vi) \((2+\sqrt{5})(3+\sqrt{2})\)
Solution:
(i) \(\sqrt{2}+2\sqrt{2}+3\sqrt{2}\)
=\(\sqrt{2}(1+2+3)\)
=\(6\sqrt{2}\)
(ii) \(5\sqrt{3}-2\sqrt{3}+\sqrt{3}\)
=\(\sqrt{3}(5-2+1)\)
=\(4\sqrt{3}\)
(iii) \((\sqrt{3}+1)(\sqrt{3}-1)\)
=\((\sqrt{3})^2-1^2\) [\(\because (a+b)(a-b)=a^2-b^2\)]
=3-1
=2
(iv) \((\sqrt{7}+3)^2\)
=\((\sqrt{7})^2+2.\sqrt{7}.3+3^2\) [\(\because (a+b)^2=a^2+2ab+b^2\)]
=\(7+6\sqrt{7}+9\)
=\(16+6\sqrt{7}\)
(v) \((2-\sqrt{5})^2\)
=\((2^2-2.2.\sqrt{5}+(\sqrt{5})^2\) [\(\because (a-b)^2=a^2-2ab+b^2\)]
=\(4-4\sqrt{5}+5\)
=\(9-4\sqrt{5}\)
(vi) (\(2+\sqrt{5})(3+\sqrt{2})\)
=\(2(3+\sqrt{2})+\sqrt{5}(3+\sqrt{2})\) [\(\because (a+b)(c+d)=a(c+d)+b(c+d)\)]
=\(6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}\)
16. হৰবিলাক পৰিমেয় কৰা।
(i) \(\LARGE \frac{1}{\sqrt{3}}\)
(ii) \(\LARGE \frac{1}{\sqrt{5}-\sqrt{3}}\)
(iii) \(\LARGE \frac{1}{\sqrt{5}-2}\)
(iv) \(\LARGE \frac{4}{\sqrt{7}+\sqrt{3}}\)
Solution:
(i) \(\LARGE \frac{1}{\sqrt{3}}\)
=\(\LARGE \frac{1×\sqrt{3}}{\sqrt{3}×\sqrt{3}}\)
=\(\LARGE \frac{\sqrt{3}}{3}\)
(ii) \(\LARGE \frac{1}{\sqrt{5}-\sqrt{3}}\)
=\(\LARGE \frac{1×(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})×(\sqrt{5}+\sqrt{3})}\)
=\(\LARGE \frac{\sqrt{5}+\sqrt{3}}{(\sqrt{5})^2-(\sqrt{3})^2}\) [\(\because (a+b)(a-b)=a^2-b^2\)]
=\(\LARGE \frac{\sqrt{5}+\sqrt{3}}{5-3}\)
=\(\LARGE \frac{\sqrt{5}+\sqrt{3}}{2}\)
(iii) \(\LARGE \frac{1}{\sqrt{5}-2}\)
=\(\LARGE \frac{1×(\sqrt{5}+2)}{(\sqrt{5}-2)×(\sqrt{5}+2)}\)
=\(\LARGE \frac{\sqrt{5}+2}{(\sqrt{5})^2-2^2}\) [\(\because (a+b)(a-b)=a^2-b^2\)]
=\(\LARGE \frac{\sqrt{5}+2}{5-4}\)
=\(\LARGE \frac{\sqrt{5}+2}{1}\)
=\(\large \sqrt{5}+2\)
(iv) \(\LARGE \frac{4}{\sqrt{7}+\sqrt{3}}\)
=\(\LARGE \frac{4×(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})×(\sqrt{7}-\sqrt{3})}\)
=\(\LARGE \frac{4×(\sqrt{7}-\sqrt{3})}{(\sqrt{7})^2-(\sqrt{3})^2}\) [\(\because (a+b)(a-b)=a^2-b^2\)]
=\(\LARGE \frac{4×(\sqrt{7}-\sqrt{3})}{7-3}\)
=\(\LARGE \frac{4×(\sqrt{7}-\sqrt{3})}{4}\)
=\(\large \sqrt{7}-\sqrt{3}\)
17. মান উলিওৱাঃ
(i) \(49^{\LARGE\frac{1}{2}}\)
(ii) \(27^{\LARGE\frac{1}{3}}\)
(iii) \(243^{\LARGE\frac{1}{5}}\)
(iv) \(125^{\LARGE\frac{2}{3}}\)
(v) \(3^{\LARGE\frac{2}{5}}.3^{\LARGE\frac{3}{5}}\)
(vi) \(64^{\LARGE\frac{-1}{3}}\)
Solution:
(i) \(49^{\LARGE\frac{1}{2}}\)
=\( (7^{\LARGE 2})^{\LARGE\frac{1}{2}} \)
=\( 7^{\LARGE{2. \frac{1}{2} }} \) [\(\because \large (a^m)^n=a^{mn} \)]
=7
(ii) \(27^{\LARGE\frac{1}{3}}\)
=\( (3^{\LARGE 3})^{\LARGE\frac{1}{3}} \)
=\( 3^{\LARGE{3. \frac{1}{3} }} \) [\(\because \large (a^m)^n=a^{mn} \)]
=3
(iii) \(243^{\LARGE\frac{1}{5}}\)
=\( (3^{\LARGE 5})^{\LARGE\frac{1}{5}} \)
=\( 3^{\LARGE{5. \frac{1}{5} }} \) [\(\because \large (a^m)^n=a^{mn} \)]
=3
(iv) \(125^{\LARGE\frac{2}{3}}\)
=\( (5^{\LARGE 3})^{\LARGE\frac{2}{3}} \)
=\( 5^{\LARGE{3. \frac{2}{3} }} \) [\(\because \large (a^m)^n=a^{mn} \)]
=\(5^2\)
=25
(v) \(3^{\LARGE\frac{2}{5}}.3^{\LARGE\frac{3}{5}}\)
=\( 3^{\LARGE\frac{2}{5}+\frac{3}{5}}\) [\(\because \large a^m.a^n=a^{m+n}\)]
=\( 3^{\LARGE \frac {2+3}{5} } \)
=\(3^{\LARGE \frac {5}{5} } \)
=3
(vi) \(64^{\LARGE\frac{-1}{3}}\)
=\( (4^{\LARGE 3})^{\LARGE\frac{-1}{3}} \)
=\( 4^{\LARGE{3. \LARGE \frac{-1}{3} }} \) [\(\because \large (a^m)^n=a^{mn} \)]
=\(4^{\LARGE -1}\)
=\(\large \frac{1}{4}\)
