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
\[{\left( {\frac{{{{\bf{x}}^{\bf{a}}}}}{{{{\bf{x}}^{\bf{b}}}}}} \right)^{{\bf{a}} + {\bf{b}}}}.{\left( {\frac{{{{\bf{x}}^{\bf{b}}}}}{{{{\bf{x}}^{\bf{c}}}}}} \right)^{{\bf{b}} + {\bf{c}}}}.{\left( {\frac{{{{\bf{x}}^{\bf{c}}}}}{{{{\bf{x}}^{\bf{a}}}}}} \right)^{{\bf{c}} + {\bf{a}}}}{\rm{ = 1}}\]
\[LHS = {\left( {\frac{{{{\bf{x}}^{\bf{a}}}}}{{{{\bf{x}}^{\bf{b}}}}}} \right)^{{\bf{a}} + {\bf{b}}}}.{\left( {\frac{{{{\bf{x}}^{\bf{b}}}}}{{{{\bf{x}}^{\bf{c}}}}}} \right)^{{\bf{b}} + {\bf{c}}}}.{\left( {\frac{{{{\bf{x}}^{\bf{c}}}}}{{{{\bf{x}}^{\bf{a}}}}}} \right)^{{\bf{c}} + {\bf{a}}}}\]
\[ = {({x^a}^{ – b})^{a{\rm{ }} + {\rm{ }}b}}.{({x^b}^{ – c})^{b{\rm{ }} + {\rm{ }}c}}.{({x^c}^{ – a})^{c{\rm{ }} + {\rm{ }}a}}\]
\[ = {x^{(a}}^{ – b)\left( {a{\rm{ }} + {\rm{ }}b} \right)}.{x^{(b}}^{ – c)\left( {b{\rm{ }} + {\rm{ }}c} \right)}.{x^{(c}}^{ – \;a)\left( {c{\rm{ }} + {\rm{ }}a} \right)}\]
\[ = {x^a}^{^2\; – {b^2}}.{x^b}^{^2\; – {c^2}}.{x^c}^{^2\; – {a^2}}\]
\[ = {x^{{a^2} – {b^2} + {b^2} – {c^2} + {c^2} – {a^2}}}\]
\[ = {\rm{ }}{x^0}\]
\[ = {\rm{ }}1\]
\[ = RHS\]
\[\therefore LHS = RHS{\rm{    }}(proved)\]
১৫। সমাধানঃ
\[{\left( {\frac{{{{\bf{x}}^{\bf{p}}}}}{{{{\bf{x}}^{\bf{q}}}}}} \right)^{p + q – r}}{\rm{ }} \times {\rm{ }}{\left( {\frac{{{x^q}}}{{{x^r}}}} \right)^{q + r – p}}{\rm{ }} \times {\rm{ }}{\left( {\frac{{{x^r}}}{{{x^p}}}} \right)^{r + p – q}}{\rm{  =  1}}\]
\[{\rm{LHS = }}{\left( {\frac{{{{\bf{x}}^{\bf{p}}}}}{{{{\bf{x}}^{\bf{q}}}}}} \right)^{p + q – r}}{\rm{ }} \times {\rm{ }}{\left( {\frac{{{x^q}}}{{{x^r}}}} \right)^{q + r – p}}{\rm{ }} \times {\rm{ }}{\left( {\frac{{{x^r}}}{{{x^p}}}} \right)^{r + p – q}}\]
\[ = {x^{\left( {p{\rm{ }} – {\rm{ }}q} \right){\rm{ }}\left( {p{\rm{ }} + {\rm{ }}q{\rm{ }} – {\rm{ }}r} \right)}}.{x^{\left( {q{\rm{ }} – {\rm{ }}r} \right){\rm{ }}\left( {q{\rm{ }} + {\rm{ }}r{\rm{ }} – {\rm{ }}p} \right)}}.{x^{\left( {r{\rm{ }} – {\rm{ }}p} \right){\rm{ }}\left( {r{\rm{ }} + {\rm{ }}p{\rm{ }} – {\rm{ }}q} \right)}}\;\]
\[ = {x^{{p^2} + pq – pr – pq – {q^2} + pq}}.{x^{{q^2} + qr – pq – qr – {r^2} + rp}}{x^{{r^2} + pr – qr – pr – {p^2} + pq}}\]
\[ = {x^{{p^2} – pr – {q^2} + pr}}.{x^{{q^2} – pq – {r^2} + rp}}{x^{{r^2} – qr – {p^2} + pq}}\]
\[ = {x^{{p^2} – pr – {q^2} + pr + {q^2} – pq – {r^2} + rp + {r^2} – qr – {p^2} + pq}}\]
\[ = {x^0}\]
\[ = {\rm{ }}1\]
\[ = RHS\]
\[\therefore LHS = RHS{\rm{    }}(proved)\]
১৬। যদি` ax = b, by = c এবং cz = a হয়, তবে দেওখাও যে xyz = 1

সমাধান : দেওয়া আছে, ax = b, by = c  এবং  cz = a

        এখন, ax = b

           বা, (cz)x = b          [∵  cz = a]

           বা , cxz = b

           বা, (by)xz= b         [∵  by = c]

            বা, bxyz = b1

           ∴ xyz = 1

[∵ ax = ay হলে  x = y যখন  a > 0, a ¹1]  (দেখানো হলো)

সমাধান করো (১৭-২০)
১৭।  4x = 8

সমাধান : 4x = 8

            বা, 2.2´x = 23
           বা, 2.2x = 23
            বা, 2x = 3               [∵ ax = ay n‡j x = y]
            ∴  x =frac{3}{2}
         নির্ণেয় সমাধান : x =frac{3}{2}

১৮। 22x + 1 = 128

সমাধান : 22x + 1 = 128

         বা, 22x.2 = 128
        বা, 22x =frac{{128}}{2}
         বা, 22x= 64
        বা, 22x= 26      [∵ 64 = 2 ´ 2 ´ 2 ´ 2 ´ 2 ´ 2 = 26]
          ∴ 2x = 6          [∵  ax= ay হলে, x = y]
         বা, x = 
            ∴ x = 3
       নির্ণেয় সমাধান : x = 3

১৯। {left( {sqrt {bf{3}} } right)^{{bf{x}} + {bf{1}}}}{rm{ = }}{left( {sqrt[{bf{3}}]{{bf{3}}}} right)^{2x - 1}}

সমাধান : ={left( {sqrt {bf{3}} } right)^{{bf{x}} + {bf{1}}}}{rm{ = }}{left( {sqrt[{bf{3}}]{{bf{3}}}} right)^{2x - 1}}

        বা, {left( {{3^{frac{1}{2}}}} right)^{x + 1}}{rm{ = }}{left( {{3^{frac{1}{3}}}} right)^{2x - 1}}    ∵ {rm{[}}{rm{ }}sqrt a = {a^{frac{1}{2}}}{rm{; }}sqrt[3]{a} = {a^{frac{1}{3}}}{rm{]}}

        বা, {{rm{3}}^{frac{1}{2} times (x + 1)}}{rm{ = }}{{rm{3}}^{frac{1}{3} times (2x - 1)}}    ∵{rm{[}}{rm{ (am}}{{rm{)}}^{rm{n}}}{rm{ = amn]}}

        বা, {{rm{3}}^{frac{{x + 1}}{2}}}{rm{ = }}{{rm{3}}^{frac{{2x - 1}}{3}}}

        ∴  frac{{x + 1}}{2} = frac{{2x - 1}}{3}                [∵ ax = ay হলে x = y]

        বা, 2(2x – 1) = 3(x + 1)    [আড়গুণণ করে]

        বা, 4x – 2 = 3x + 3

        বা, 4x – 3x = 3 + 2 

        ∴x = 5

       নির্ণেয় সমাধান : x = 5

২০। 2x + 21– x = 3

সমাধান : 2x + 21– x = 3

         বা, 2x + 2.2 – x = 3
         বা, 2x(2x + 2.2 – x) = 3×2x        [উভয় পক্ষকে 2x  দ্বারা গুণ করে]
          বা, 2x+x + 2.2 – x + x = 3 × 2x
          বা, 22x + 2.20 = 3× 2x
          বা, 22x + 2.1 = 3 × 2x
          বা, (2x)2 + 2 = 3 × 2x
          বা, (2x)2 – 3× 2x + 2 = 0
         বা, a2 – 3a + 2 = 0                     [2x= a ধরে]
          বা, a(a – 2) – 1(a – 2) = 0
          বা, (a – 2) (a – 1) = 0
        হয়, a – 2 = 0                             অথবা, a – 1 = 0
         বা, a = 2                                     বা, a = 1
         বা, 2x = 2 [মান বসিয়ে]      বা, 2x = 1 [মান বসিয়ে]
          বা, 2x = 21                                 বা, 2x = 20         [∵ 20 =1]
         ∴ x = 1                                      ∴ x = 0
        নির্ণেয় সমাধান: x = 0, 1

২১. P=xa, Q=xb এবং R=xc

ক) Pbc.Q-ca = এর মান নির্ণয় কর।

খ) {left( {frac{Q}{P}} right)^{a + b}} times {left( {frac{Q}{R}} right)^{b + c}} div 2{left( {frac{R}{P}} right)^{a - c}}এর মান নির্ণয় কর।

গ) দেখাও যে, {left( {frac{P}{Q}} right)^a}{^{^2}^{ + ab + b}}^{^2} times {left( {frac{Q}{R}} right)^b}{^{^2}^{ + bc + c}}^{^2} times {left( {frac{R}{P}} right)^c}{^{^2}^{ + ca + a}}^{^2} = 1

সমাধানঃ

ক)
Pbc.Q-ca
=(xa)bc.(xb)-ca
=xabc.x-abc
=xabc-abc
=x0
=1

খ) = {left( {frac{{{x^a}}}{{{x^b}}}} right)^{a + b}} times {left( {frac{{{x^b}}}{{{x^c}}}} right)^{b + c}} div 2{left( {frac{{{x^c}}}{{{x^a}}}} right)^{a - c}}

= {left( {{x^{a - b}}} right)^{a + b}} times {left( {{x^{b - c}}} right)^{b + c}} div 2{left( {{x^{c - a}}} right)^{a - c}}

= {x^{{a^2} - {b^2}}}.{x^{{b^2} - {c^2}}}.frac{1}{{2{x^{{c^2} - {a^2}}}}}

= {x^{{a^2} - {b^2} + {b^2} - {c^2} - {c^2} + {a^2}}}frac{1}{2}

= {x^0}frac{1}{2}

= 1 times frac{1}{2}

= frac{1}{2}

গ) LHS = {left( {frac{P}{Q}} right)^a}{^{^2}^{ + ab + b}}^{^2} times {left( {frac{Q}{R}} right)^b}{^{^2}^{ + bc + c}}^{^2} times {left( {frac{R}{P}} right)^c}{^{^2}^{ + ca + a}}^{^2}
= {left( {{x^{a - b}}} right)^a}{^{^2}^{ + ab + b}}^{^2} times {left( {{x^{b - c}}} right)^b}{^{^2}^{ + bc + c}}^{^2} times {left( {{x^{c - a}}} right)^c}{^{^2}^{ + ca + a}}^{^2}
= {x^{(a - b)({a^2} + ab + {b^2})}} times {x^{(b - c)({b^2} + bc + {c^2})}} times {x^{(c - a)({c^2} + ac + {a^2})}}
= {x^{{a^3} - {b^3}}} times {x^{{b^3} - {c^3}}} times {x^{{c^3} - {a^3}}}
= {x^{{a^3} - {b^3} + {b^3} - {c^3} + {c^3} - {a^3}}}
= {x^0}
= 1
= RHS
∴LHS=RHS (proved)
২২। X = {(2{a^{ - 1}} + 3{b^{ - 1}})^{ - 1}} ,Y = sqrt[{pq}]{{frac{{{x^p}}}{{{x^q}}}}} times sqrt[{qr}]{{frac{{{x^q}}}{{{x^r}}}}} times sqrt[{rp}]{{frac{{{x^r}}}{{{x^p}}}}}
এবং Z = frac{{{5^{m + 1}}}}{{{{({5^m})}^{m - 1}}}} div frac{{{{25}^{m + 1}}}}{{{{({5^{m - 1}})}^{m + 1}}}} যেখানে p,q,r,s＞0
ক) x এর মান নির্ণয় কর।
খ) দেখাও যে Y + sqrt[4]{{81}} = 4
গ) দেখাও যে Y div Z = 25
সমাধানঃ
ক) begin{array}{ccccccccccccccc} x&{ = {rm{ }}{{left( {2{a^{ - 1}} + 3{b^{ - 1}}} right)}^{ - 1}}} end{array}
    = {(2 times frac{1}{a} + 3 times frac{1}{b})^{ - 1}}
= {(frac{2}{a} + frac{3}{b})^{ - 1}}
= {(frac{{2b + 3a}}{{ab}})^{ - 1}}
= frac{{ab}}{{2b + 3a}}
খ) দেওয়া আছে, Y = sqrt[{pq}]{{frac{{{x^p}}}{{{x^q}}}}} times sqrt[{qr}]{{frac{{{x^q}}}{{{x^r}}}}} times sqrt[{rp}]{{frac{{{x^r}}}{{{x^p}}}}}
এখন, LHS=  Y + sqrt[4]{{81}}
= sqrt[{pq}]{{frac{{{x^p}}}{{{x^q}}}}} times sqrt[{qr}]{{frac{{{x^q}}}{{{x^r}}}}} times sqrt[{rp}]{{frac{{{x^r}}}{{{x^p}}}}} + sqrt[4]{{81}}
\[ = {\left( {\frac{{{x^p}}}{{{x^q}}}} \right)^{\frac{1}{{pq}}}} \times {\left( {\frac{{{x^q}}}{{{x^r}}}} \right)^{\frac{1}{{qr}}}} \times {\left( {\frac{{{x^r}}}{{{x^p}}}} \right)^{\frac{1}{{rp}}}} + {(81)^{\frac{1}{4}}}\]
= {left( {{x^{p - q}}} right)^{frac{1}{{pq}}}} times {left( {{x^{q - r}}} right)^{frac{1}{{qr}}}} times {left( {{x^{r - p}}} right)^{frac{1}{{rp}}}} + {({3^4})^{frac{1}{4}}}
= {x^{frac{{p - q}}{{pq}}}} times {x^{frac{{q - r}}{{qr}}}} times {x^{frac{{r - p}}{{rp}}}} + {3^{4 times frac{1}{4}}}
= {x^{frac{{p - q}}{{pq}} + frac{{q - r}}{{qr}} + frac{{r - p}}{{rp}}}} + {3^1}
= {x^{frac{{pr - qr + qp - rp + rq - rp}}{{pqr}}}} + 3
= {x^{frac{0}{{pqr}}}} + 3
= {x^0} + 3
= 1 + 3
= 4
= RHS
∴LHS=RHS (proved)
গ) দেওয়া আছে, Z = frac{{{5^{m + 1}}}}{{{{({5^m})}^{m - 1}}}} div frac{{{{25}^{m + 1}}}}{{{{({5^{m - 1}})}^{m + 1}}}}
= frac{{{5^{m + 1}}}}{{{5^m}^{^2 - m}}} div frac{{{{({5^2})}^{m + 1}}}}{{{5^{{m^2} - {1^2}}}}}
= frac{{{5^{m + 1}}}}{{{5^m}^{^2 - m}}} div frac{{{5^2}^{m + 2}}}{{{5^{{m^2} - 1}}}}
= {5^{m + 1 - {m^2} + m}} div {5^2}^{m + 2 - {m^2} + 1}
= {5^{2m + 1 - {m^2}}} div {5^2}^{m + 3 - {m^2}}
= {5^{2m + 1 - {m^2} - 2m - 3 + {m^2}}}
= {5^{ - 2}}
= frac{1}{{{5^2}}}
= frac{1}{{25}}
খ থেকে পাই Y=1
 এখন, Y div Z
= 1 div frac{1}{{25}}
= 1 times frac{{25}}{1}
= 25
∴Y div Z = 25 (দেখানো হলো)
———– সমাপ্ত———–

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