Question

can someone help me with this sum with steps.. ​

Answer 1

GIVEN :-

$\implies \sf \: \dfrac{x ^{2n + 3} . \: x ^{(2n +1)(n + 2) } }{(x^{3})^{2n + 1} \: . \: x ^{n(2n + 1)} }$

TO FIND :-

$\implies \sf value \: of \: \: \dfrac{x ^{2n + 3} . \: x ^{(2n +1)(n + 2) } }{(x^{3})^{2n + 1} \: . \: x ^{n(2n + 1)} }$

SOLUTION :-

$\to \sf \: \dfrac{x ^{2n + 3} . \: x ^{(2n +1)(n + 2) } }{(x^{3})^{2n + 1} \: . \: x ^{n(2n + 1)} } \\ \\ \to \: \sf \dfrac{x ^{2n + 3} \: . \: x ^{2n(n + 2) + 1(n + 2)} }{x ^{6n + 3} \: . \: x ^{2n ^{2} + n} } \\ \\ \to \sf \: \dfrac{x ^{2n + 3 } \: . \: x ^{2n ^{2} + 4n + n + 2} }{x ^{6n + 3} \: . \: x ^{2n^{2} + n} }$

➠ By using identity: x² . x³ = x² + ³.

$\to \sf \: \dfrac{x ^{(2n + 3) + (2n^{2} + 4n +n + 2)} }{x ^{(6n + 3 )+ (2n ^{2} + n) } } \\ \\ \to \sf \: \dfrac{x ^{2n + 3 + 2n^{2} + 4n +n + 2} }{x ^{6n + 3 + 2n ^{2} + n} } \\ \\ \to \sf \: \dfrac{x ^{2n ^{2} + 7n +5 } }{x ^{2n ^{2} + 7n + 3 } }$

➠ By using identity: x² ÷ x³ = x²-³

$\to \sf \: x ^{(2n ^{2 } + 7n + 5) - (2n ^{2} + 7n + 3)} \\ \\ \to \sf \: x ^{ \cancel{2n ^{2} } + \cancel{7n} + 5 \cancel{-2n ^{2}} \cancel{ - 7n} - 3 } \\ \\ \to \sf \: x ^{5 - 3} \\ \\ \to \boxed{ \red{ \bf \: x ^{2} }}$

Hence the required answer is x².

ADDITIONAL INFORMATION :-

$\boxed{\begin{minipage}{7cm} \\ \sf{ \implies \bf \sqrt[n]{ \sqrt[m]{ \sqrt[p]{((a^{x} )^{y}) ^{z} } } } = (a ^{xyz} )^{ \frac{1}{mnp} } = a ^{ \frac{xyz}{mnp} } } \\ \\ \sf{ \implies a^m \times a^n = a^{m+n} } \\ \\ \sf{ \implies {a}^{m} \times b^m = ab^m } \\ \\ \sf{ \implies \dfrac{a^m}{a^n} = a^{m - n} ( \tt{ If \: m > n} ) } \\ \\ \sf{ \implies \dfrac{a^m}{ a^n} = \dfrac{ 1}{ a^{n-m} } ( \tt{ If \: n > m )} } \\ \\ \sf{ \implies (a^m)^n = a^{mn} } \\ \\ \sf{ \implies a^{-n} = \dfrac{1}{ a^n} }\end{minipage}}$

Answer 2

idk.

sry for it!!!

“Get lost, creep.”❤️