$\left( \frac{3^{m+1}}{(3^m)^{m-1}} \div \frac{9^{m+1}}{(3^{m-1})^{m+1}} \right) \times 3$ = কত ?
Solution
Correct Answer: Option C
সূচকের অংশটি যদি আমরা সরল করি:
$\frac{{{3^{m + 1}}}}{{{{({3^m})}^{m - 1}}}} \div \frac{{{9^{m + 1}}}}{{{{({3^{m - 1}})}^{m + 1}}}}$
= $\frac{{{3^{m + 1}}}}{{{3^{m(m - 1)}}}} \div \frac{{{({3^2})}^{m + 1}}}{{{3^{(m - 1)(m + 1)}}}}$
= $\frac{{{3^{m + 1}}}}{{{3^{{m^2} - m}}}} \div \frac{{{3^{2m + 2}}}}{{{3^{{m^2} - 1}}}}$
= ${3^{(m + 1) - ({m^2} - m)}} \div {3^{(2m + 2) - ({m^2} - 1)}}$
= ${3^{m + 1 - {m^2} + m}} \div {3^{2m + 2 - {m^2} + 1}}$
= ${3^{2m - {m^2} + 1}} \div {3^{2m - {m^2} + 3}}$
= $\frac{{{3^{2m - {m^2} + 1}}}}{{{3^{2m - {m^2} + 3}}}}$
= ${3^{(2m - {m^2} + 1) - (2m - {m^2} + 3)}}$
= ${3^{2m - {m^2} + 1 - 2m + {m^2} - 3}}$
= ${3^{1 - 3}}$
= ${3^{ - 2}}$
= $\frac{1}{{{3^2}}}$
= $\frac{1}{9}$
$\left( \frac{3^{m+1}}{(3^m)^{m-1}} \div \frac{9^{m+1}}{(3^{m-1})^{m+1}} \right) \times 3$
তাহলে: $\frac{1}{9} \times 3 = \frac{1}{3}$
সুতরাং, সঠিক উত্তর: 1/3
$\frac{{{3^{m + 1}}}}{{{{({3^m})}^{m - 1}}}} \div \frac{{{9^{m + 1}}}}{{{{({3^{m - 1}})}^{m + 1}}}}$
= $\frac{{{3^{m + 1}}}}{{{3^{m(m - 1)}}}} \div \frac{{{({3^2})}^{m + 1}}}{{{3^{(m - 1)(m + 1)}}}}$
= $\frac{{{3^{m + 1}}}}{{{3^{{m^2} - m}}}} \div \frac{{{3^{2m + 2}}}}{{{3^{{m^2} - 1}}}}$
= ${3^{(m + 1) - ({m^2} - m)}} \div {3^{(2m + 2) - ({m^2} - 1)}}$
= ${3^{m + 1 - {m^2} + m}} \div {3^{2m + 2 - {m^2} + 1}}$
= ${3^{2m - {m^2} + 1}} \div {3^{2m - {m^2} + 3}}$
= $\frac{{{3^{2m - {m^2} + 1}}}}{{{3^{2m - {m^2} + 3}}}}$
= ${3^{(2m - {m^2} + 1) - (2m - {m^2} + 3)}}$
= ${3^{2m - {m^2} + 1 - 2m + {m^2} - 3}}$
= ${3^{1 - 3}}$
= ${3^{ - 2}}$
= $\frac{1}{{{3^2}}}$
= $\frac{1}{9}$
$\left( \frac{3^{m+1}}{(3^m)^{m-1}} \div \frac{9^{m+1}}{(3^{m-1})^{m+1}} \right) \times 3$
তাহলে: $\frac{1}{9} \times 3 = \frac{1}{3}$
সুতরাং, সঠিক উত্তর: 1/3
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