Question

Pls help me answer this question thx

Answer 1

Answer:

$\large\boxed{\dfrac{1}{115}}$

Step-by-step explanation:

$\text{Use}\\\\(a^n)^m=a^{nm}\\\\a^n\cdot a^m=a^{n+m}\\\\(a\cdot b)^n=a^n\cdot b^n\\\\a^{-1}=\dfrac{1}{a}\\--------------------------\\\\\dfrac{5^{2n-1}-4(25^{n-1})}{(5^n\cdot2)^2+3(5^{2n-1})}=\dfrac{5^{2n}\cdot5^{-1}-4\left[(5^2)^n\cdot25^{-1}\right]}{(5^n)^2\cdot2^2+3\left[(5^2)^n\cdot5^{-1}\right]}$

$=\dfrac{5^{2n}\cdot\frac{1}{5}-4\left(5^{2n}\cdot\frac{1}{25}\right)}{5^{2n}\cdot4+3\left(5^{2n}\cdot\frac{1}{5}\right)}=\dfrac{5^{2n}\left(\frac{1}{5}-\frac{4}{25}\right)}{5^{2n}\left(4+\frac{3}{5}\right)}=\dfrac{\frac{5}{25}-\frac{4}{25}}{4\frac{3}{5}}=\dfrac{1}{25}:\left(\dfrac{4\cdot5+3}{5}\right)\\\\=\dfrac{1}{25}\cdot\dfrac{5}{23}=\dfrac{1}{5}\cdot\dfrac{1}{23}=\dfrac{1}{115}$