নবম-দশম শ্রেণির গণিত অনুশীলনী ৪.১ সূচক, সরল, প্রমাণ ও সমাধান SSC Math Chapter 4.1 Exponents
নবম-দশম শ্রেণির গণিত অনুশীলনী ৪.১ সূচক, সরল, প্রমাণ ও সমাধান SSC Math Chapter 4.1 Exponents পোস্টে সকলকে স্বাগতম। আজকে আমরা এসএসসি বা নবম-দশম শ্রেণীর অনুশীলনী ৪.১ সূচক এর গণিত বইয়ের প্রশ্নগুলোর উত্তর জানবো। গণিত চতুর্থ অধ্যায় অনুশীলনী 4.1 শিক্ষার্থীদের কাছে অনেক সহজ মনে হতে পারে। তবে শিক্ষার্থীরা এখানে ছোট ছোট ভুল করে থাকে ফলে সমস্যাগুলো সমাধান ভুল হয়ে যায়। তাই শিক্ষার্থীদের বলব তোমরা এই অধ্যায়টি খুব মনোযোগ সহকারে করবে। নবম-দশম গণিত অনুশীলনী ৪.১ সূচক প্রশ্ন উত্তর সরল কর (১-৮) ১। সমাধানঃ \[\frac{{{7^3} \times {7^{ – 3}}}}{{3 \times {3^{ – 4}}}}\] \[ = \frac{{{7^3}^{ – 3}}}{{{3^{1 – 4}}}}\] \[ = \frac{{{7^0}}}{{{3^{ – 3}}}}\] \[ = \frac{1}{{\frac{1}{{{3^3}}}}}\] \[ = \frac{1}{{\frac{1}{{27}}}}\] \[ = 27\] ২। সমাধানঃ \[ = \frac{{\sqrt[3]{{{7^2}}}.\sqrt[3]{7}}}{{\sqrt 7 }}\] \[ = \frac{{{7^{\frac{2}{3}}}{{.7}^{\frac{1}{3}}}}}{{{7^{\frac{1}{2}}}}}\] \[ = \frac{{{7^{\frac{2}{3} + }}^{\frac{1}{3}}}}{{{7^{\frac{1}{2}}}}}\] \[ = \frac{{{7^{\frac{{2 + 1}}{3}}}}}{{{7^{\frac{1}{2}}}}}\] \[ = \frac{{{7^{\frac{3}{3}}}}}{{{7^{\frac{1}{2}}}}}\] \[ = \frac{{{7^1}}}{{{7^{\frac{1}{2}}}}}\] \[ = {7^{1 – \frac{1}{2}}}\] \[ = {7^{\frac{{2 – 1}}{2}}}\] \[ = {7^{\frac{1}{2}}}\] \[ = \sqrt 7 \] ৩। সমাধানঃ \[ = {({2^{ – 1}} + {5^{ – 1}})^{ – 1}}\] \[ = {(\frac{1}{2} + \frac{1}{5})^{ – 1}}\] \[ = {(\frac{5}{{10}} + \frac{2}{5})^{ – 1}}\] \[ = {(\frac{7}{{10}})^{ – 1}}\] \[ = \frac{1}{{\frac{7}{{10}}}}\] \[ = \frac{{10}}{7}\] ৪। সমাধানঃ \[{ = {{\left( {{\bf{2}}{{\bf{a}}^{ – {\bf{1}}}} + {\bf{3}}{{\bf{b}}^{ – {\bf{1}}}}} \right)}^{ – {\bf{1}}}}}\] \[ = {\left( {{\bf{2}}\frac{1}{a} + {\bf{3}}\frac{1}{b}} \right)^{ – {\bf{1}}}}\] \[ = {\left( {\frac{2}{a} + \frac{3}{b}} \right)^{ – {\bf{1}}}}\] \[ = {\left( {\frac{{2b + 3a}}{{ab}}} \right)^{ – {\bf{1}}}}\] \[ = \frac{{ab}}{{3a + 2b}}\] ৫। সমাধানঃ \[ = {(\frac{{{{\bf{a}}^{\bf{2}}}{{\bf{b}}^{ – {\bf{1}}}}}}{{{{\bf{a}}^{ – {\bf{1}}}}{\bf{b}}}})^{\bf{2}}}\] \[ = {(\frac{{{{\bf{a}}^{\bf{2}}}\frac{1}{b}}}{{\frac{1}{{{a^2}}}{\bf{b}}}})^{\bf{2}}}\] \[ = {(\frac{{\frac{{{{\bf{a}}^{\bf{2}}}}}{b}}}{{\frac{{\bf{b}}}{{{a^2}}}}})^{\bf{2}}}\] \[ = {(\frac{{{a^2}.{a^2}}}{{b.b}})^{\bf{2}}}\] \[ = {(\frac{{{a^4}}}{{{b^2}}})^{\bf{2}}}\] \[ = \frac{{{a^8}}}{{{b^4}}}\] ৬। সমাধানঃ \[ = \sqrt {{x^{ – 1}}y} .\sqrt {{y^{ – 1}}z} .\sqrt {{z^{ – 1}}x} {\rm{ }}(x > 0,y > 0,z > 0)\] \[ = \sqrt {\frac{1}{x}y} .\sqrt {\frac{1}{y}z} .\sqrt {\frac{1}{z}x} \] \[ = \sqrt {\frac{y}{x}} .\sqrt {\frac{z}{y}} .\sqrt {\frac{x}{z}} \] \[ = \sqrt {\frac{{y \times z \times x}}{{x \times y \times z}}} \] \[ = \sqrt 1 \] \[ = 1\] ৭। সমাধানঃ \[ = \frac{{{2^{n + 4}} – {{4.2}^{n + 1}}}}{{{2^{n + 2}} \div 2}}\] \[ = \frac{{{2^n}{2^4} – {{4.2}^n}{2^1}}}{{{2^n}{2^2} \div {2^1}}}\] \[ = \frac{{{2^n}({2^4} – {{4.2}^1})}}{{{2^n}{2^{2 – 1}}}}\] \[ = \frac{{(16 – 8)}}{{{2^1}}}\] \[ = \frac{8}{2}\] \[ = 4\] ৮। সমাধানঃ \[ = \frac{{{3^{m + 1}}}}{{{{({3^m})}^{m – 1}}}} \div \frac{{{9^{m + 1}}}}{{{{({3^{m – 1}})}^{m + 1}}}}\] \[ = \frac{{{3^{m + 1}}}}{{{3^{{m^2} – m}}}} \div \frac{{{{({3^2})}^{m + 1}}}}{{{3^{{m^2} – 1}}}}\] \[ = \frac{{{3^{m + 1}}}}{{{3^{{m^2} – m}}}} \div \frac{{{3^2}^{m + 2}}}{{{3^{{m^2} – 1}}}}\] \[ = {3^{m + 1 – {m^2} + m}} \div {3^{2m + 2 – {m^2} + 1}}\] \[ = {3^{2m + 1 – {m^2}}} \div {3^{2m + 3 – {m^2}}}\] \[ = {3^{2m + 1 – {m^2} – }}^{2m – 3 + {m^2}}\] \[ = {3^{ – 2}}\] \[ = \frac{1}{{{3^2}}}\] \[ = \frac{1}{9}\] প্রমাণ করো (৯-১৫) ৯। সমাধানঃ \[frac{{{4^n} – 1}}{{{2^n} – 1}} = \frac{{{{({2^2})}^n} – 1}}{{{2^n} – 1}}\] \[LHS = \frac{{{4^n} – 1}}{{{2^n} – 1}}\] \[ = \frac{{{{({2^2})}^n} – 1}}{{{2^n} – 1}}\] \[ = \frac{{{2^2}^n – 1}}{{{2^n} – 1}}\] \[ = \frac{{({2^2}^n – 1)({2^2}^n + 1)}}{{({2^n} – 1)({2^2}^n + 1)}}\] \[ = \frac{{({2^2}^n – 1)({2^2}^n + 1)}}{{{{({2^n})}^2} – {1^2}}}\] \[ = \frac{{({2^2}^n – 1)({2^2}^n + 1)}}{{({2^{2n}} – 1)}}\] \[ = {2^2}^n + 1\] \[ = RHS\] \[\therefore LHS = RHS{\rm{ (proved)}}\] ১০। সমাধানঃ \[\frac{{{2^{2p + 1}}{{.3}^{2p + q}}{{.5}^{p + q}}{{.6}^p}}}{{{3^{p – 2}}{{.6}^{2p + 2}}{{.10}^p}{{.15}^q}}} = \frac{1}{2}\] \[LHS = \frac{{{2^{2p + 1}}{{.3}^{2p + q}}{{.5}^{p + q}}{{.6}^p}}}{{{3^{p – 2}}{{.6}^{2p + 2}}{{.10}^p}{{.15}^q}}}\] \[ = \frac{{{2^{2p + 1}}{{.3}^{2p + q}}{{.5}^{p + q}}.{{(2 \times 3)}^p}}}{{{3^{p – 2}}.{{(2 \times 3)}^{2p + 2}}.{{(2 \times 5)}^p}.{{(3 \times 5)}^q}}}\] \[ = \frac{{{2^{2p + 1}}{{.3}^{2p + q}}{{.5}^{p + q}}{{.2}^p}{{.3}^p}}}{{{3^{p – 2}}{{.2}^{2p + 2}}{{.3}^{2p + 2}}{{.2}^p}{{.5}^p}{{.3}^q}{{.5}^q}}}\] \[ = \frac{{{2^{2p + 1 + p}}{{.3}^{2p + q + p}}{{.5}^{p + q}}}}{{{3^{p – 2 + }}^{2p + 2 + q}{{.2}^{2p + 2 + p}}{{.5}^{p + }}^q}}\] \[ = \frac{{{2^{3p + 1}}{{.3}^{3p + q}}{{.5}^{p + q}}}}{{{3^{3p + q}}{{.2}^{3p + 2}}{{.5}^{p + }}^q}}\] \[ = {2^{3p + 1 – 3p – 2}}{.3^{3p + q – 3p – q}}{.5^{p + q – p – q}}\] \[ = {2^{ – 1}}{.3^0}{.5^0}\] \[ = {2^{ – 1}}.1.1\] \[ = \frac{1}{2}\] \[ = RHS\] \[\therefore LHS = RHS(proved)\] ১১। সমাধানঃ \[{(\frac{{{a^l}}}{{{a^m}}})^n}.{(\frac{{{a^m}}}{{{a^n}}})^l}{(\frac{{{a^n}}}{{{a^l}}})^m} = 1\] \[LHS = {(\frac{{{a^l}}}{{{a^m}}})^n}.{(\frac{{{a^m}}}{{{a^n}}})^l}{(\frac{{{a^n}}}{{{a^l}}})^m}\] \[ = {a^{\ln – m}}^n.{a^{ml – n}}^l{a^{nm – l}}^m\] \[ = {a^{\ln – mn + ml – nl + nm – lm}}\] \[ = {a^0}\] \[ = 1\] \[ = RHS\] \[\therefore LHS = RHS{\rm{ }}(proved)\] ১২। সমাধানঃ \[\frac{{{a^{p + q}}}}{{{a^{2r}}}} \times \frac{{{a^{q + q}}}}{{{a^{2p}}}} \times \frac{{{a^{r + p}}}}{{{a^{2q}}}} = 1\] \[LHS = \frac{{{a^{p + q}}}}{{{a^{2r}}}} \times \frac{{{a^{q + q}}}}{{{a^{2p}}}} \times \frac{{{a^{r + p}}}}{{{a^{2q}}}}\] \[ = \frac{{{a^{p + q – 2r + p + q + r + p}}}}{{{a^{2r + 2p + 2q}}}}\] \[ = \frac{{{a^{2p + 2q + 2r}}}}{{{a^{2r + 2p + 2q}}}}\] \[ = {a^{2p + 2q + 2r – 2p – 2q – 2r}}\] \[ = {a^0}\] \[ = 1\] \[ = RHS\] \[\therefore LHS = RHS{\rm{ }}(proved)\] ১৩। সমাধানঃ \[{(\frac{{{x^a}}}{{{a^b}}})^{\frac{1}{{ab}}}}.{(\frac{{{x^b}}}{{{a^c}}})^{\frac{1}{{bc}}}}.{(\frac{{{x^c}}}{{{a^a}}})^{\frac{1}{{ca}}}} = 1\] \[LHS = {(\frac{{{x^a}}}{{{a^b}}})^{\frac{1}{{ab}}}}.{(\frac{{{x^b}}}{{{a^c}}})^{\frac{1}{{bc}}}}.{(\frac{{{x^c}}}{{{a^a}}})^{\frac{1}{{ca}}}}\] \[ = {({x^{a – b}})^{\frac{1}{{ab}}}}.{({x^{b – c}})^{\frac{1}{{bc}}}}.{({x^{c – a}})^{\frac{1}{{ca}}}}\] \[ = {x^{\frac{{a – b}}{{ab}}}}.{x^{\frac{{b – c}}{{bc}}}}.{x^{\frac{{c – a}}{{ca}}}}\] \[ = {x^{\frac{{a – b}}{{ab}} + \frac{{b – c}}{{bc}} + \frac{{c – a}}{{ca}}}}\] \[ = {x^{\frac{{c(a – b) + a(b – c) + b(c – a)}}{{abc}}}}\] \[ = {x^{\frac{{ac – bc + ab – ac + bc – ab}}{{abc}}}}\] \[ = {x^{\frac{0}{{abc}}}}\] \[ = {x^0}\] \[ = 1\] \[ = RHS\] \[\therefore LHS = RHS{\rm{ }}(proved)\] ১৪। সমাধানঃ \[{\left( {\frac{{{{\bf{x}}^{\bf{a}}}}}{{{{\bf{x}}^{\bf{b}}}}}} \right)^{{\bf{a}} + {\bf{b}}}}.{\left( {\frac{{{{\bf{x}}^{\bf{b}}}}}{{{{\bf{x}}^{\bf{c}}}}}} \right)^{{\bf{b}} + {\bf{c}}}}.{\left( {\frac{{{{\bf{x}}^{\bf{c}}}}}{{{{\bf{x}}^{\bf{a}}}}}} \right)^{{\bf{c}} + {\bf{a}}}}{\rm{ = 1}}\] \[LHS = {\left( {\frac{{{{\bf{x}}^{\bf{a}}}}}{{{{\bf{x}}^{\bf{b}}}}}} \right)^{{\bf{a}} + {\bf{b}}}}.{\left( {\frac{{{{\bf{x}}^{\bf{b}}}}}{{{{\bf{x}}^{\bf{c}}}}}} \right)^{{\bf{b}} + {\bf{c}}}}.{\left( {\frac{{{{\bf{x}}^{\bf{c}}}}}{{{{\bf{x}}^{\bf{a}}}}}} \right)^{{\bf{c}} + {\bf{a}}}}\] \[ = {({x^a}^{ – b})^{a{\rm{ }} + {\rm{ }}b}}.{({x^b}^{ – c})^{b{\rm{ }} + {\rm{ }}c}}.{({x^c}^{ – a})^{c{\rm{ }} + {\rm{ }}a}}\] \[ = {x^{(a}}^{ – b)\left( {a{\rm{ }} + {\rm{ }}b} \right)}.{x^{(b}}^{ – c)\left( {b{\rm{ }} + {\rm{ }}c} \right)}.{x^{(c}}^{ – \;a)\left( {c{\rm{ }} + {\rm{ }}a} \right)}\] \[ = {x^a}^{^2\; – {b^2}}.{x^b}^{^2\; – {c^2}}.{x^c}^{^2\; – {a^2}}\] \[ = {x^{{a^2} – {b^2} + {b^2} – {c^2}