Question

9th Grade Maths. Solutions please
It would be helpful
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Answer 1

Answer:

Note :

$= > \: \: \: {a}^{m} \times {a}^{n} = {a}^{m + n} \\ \\ = > \: \: \: {a}^{m} = {a}^{n} \: \: \: \: \therefore \: \: \: m = n$

Now,

$\sf \: \: \: \: \: \: \: \frac{ {a}^{m}. {a}^{n} }{ {a}^{k} } = {a}^{p} \\ \sf \implies \: {a}^{m} . {a}^{n} = {a}^{p} . {a}^{k} \\ \sf \implies \: {a}^{m + n} = {a}^{p + k} \\ \sf \therefore \: \: \: \: \: \: \: m + n = p + k$

So option (d) is the correct answer.

Answer 2

$\large\underline{\sf{Solution-}}$

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:\dfrac{ {a}^{m} \: . \: {a}^{n} }{ {a}^{k} } = {a}^{p}$

We know,

$\boxed{ \bf{ \: {x}^{m} \times {x}^{n} = {x}^{m + n}}}$

So, using this identity in numerator, we get

$\rm :\longmapsto\:\dfrac{ {a}^{m + n} }{ {a}^{k} } = {a}^{p}$

We know that,

$\boxed{ \bf{ \: {x}^{m} \div {x}^{n} = {x}^{m - n}}}$

So using this identity, we get

$\rm :\longmapsto\: {a}^{m + n - k} = {a}^{p}$

Now, we know that,

$\boxed{ \bf{ \: {x}^{m} = {x}^{n} \: \implies \: m = n}}$

So, using this result we get

$\rm :\longmapsto\:m + n - k = p$

can be rewritten as

$\rm :\longmapsto\:m + n = p + k$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \red{ \underbrace{ \boxed{ \bf{ \: \: \: Option \: D) \: is \: correct \: \: \: }}}}$

Additional Information :-

Important identities :-

$\boxed{ \bf{ \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}}$

$\boxed{ \bf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\boxed{ \bf{ \: {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}$

$\boxed{ \bf{ \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

$\boxed{ \bf{ \: {x}^{2} - {y}^{2} = (x - y)(x + y)}}$

$\boxed{ \bf{ \: {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}}$

$\boxed{ \bf{ \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} )}}$

$\boxed{ \bf{ \: {x}^{4} - {y}^{4} = (x - y)(x + y)( {x}^{2} + {y}^{2} )}}$