Question

Simplify:
$\frac{3^{30}+3^{29}+3^{25}}{3^{31}+3^{30}-3^{29}}+\frac{2^{30}+2^{29}+2^{25}}{2^{31}+2^{30}-2^{29}}$

Answer 1

We have,

$=\frac{3^{30}+3^{29}+3^{25}}{3^{31}+3^{30}-3^{29}}+\frac{2^{30}+2^{29}+2^{25}}{2^{31}+2^{30}-2^{29}}$

$=\frac{3^{25}(3^5+3^4+1)}{3^{29}(3^2+3-1)}+\frac{2^{25}(2^5+2^4+1)}{2^{29}(2^2+2-1)}$

$=\frac{(243+81+1)}{3^4(9+3-1)}+\frac{(32+16+1)}{2^4(4+2-1)}$

$=\frac{325}{3^4\times 11}+\frac{49}{2^4\times 5}$

$=\frac{325}{81\times 11}+\frac{49}{16\times 5}$

$=\frac{325}{891}+\frac{49}{80}$

$=\frac{325\times 80+49\times 891}{891\times 80}$

$=\frac{26000+43659}{71280}$

$=\frac{69659}{71280}$

Hence, the value is $\frac{69659}{71280}$.