Lời giải:

Xét tử số:

$\text{TS}=1+25^4+25^8+...+25^{28}$

$25^4.\text{TS}=25^4+25^8+...+25^{32}$

$\Rightarrow \text{TS}(25^4-1)=25^{32}-1$

$\text{TS}=\frac{25^{32}-1}{25^4-1}$

Xét mẫu số:

$\text{MS}=1+25^2+..+25^{30}$

$25^2.\text{MS}=25^2+25^4+...+25^{32}$

$\Rightarrow \text{MS}(25^2-1)=25^{32}-1$

$\Rightarrow \text{MS}=\frac{25^{32}-1}{25^2-1}$

Do đó:
$A=\frac{25^{32}-1}{25^4-1}:\frac{25^{32}-1}{25^2-1}=\frac{25^2-1}{25^4-1}$

$=\frac{25^2-1}{(25^2-1)(25^2+1)}=\frac{1}{25^2+1}$

\(A=\frac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+25^{26}+...+25^2+1}=25^{30}+25^{26}+25^{22}+25^{18}+25^{14}+25^{10}+25^6+25^2\)

a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)

\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)

\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)

\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)

\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)

\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)

\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)

\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)