a) A=\(\frac{2^{13}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^5-25^3.49^2}{\left(125.7\right)^3+5^9.14^3}\) =\(\frac{2^{13}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^5-\left(5^2\right)^3.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(2.7\right)^3}\) =\(\frac{2^{13}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^5-5^6.7^4}{5^9.7^3+5^9.2^3.7^3}\) =\(\frac{2^{12}.3^4.\left(2.3-1\right)}{2^{12}.3^5.\left(6+1\right)}-\frac{5^6.7^4.\left(5^4.7-1\right)}{2^3.5^9.7^3\left(1+1.2^3\right)}\) =\(\frac{2^{12}.3^4.5}{2^{12}.3^5.7}-\frac{5^6.7^4.4374}{3^3.5^9.7^3.9}\) =\(\frac{5}{3.7}-\frac{7.4374}{3^3.5^3.3^2}\) =\(\frac{5}{21}-\frac{7.4374}{3^6.5^3}\) =\(\frac{5}{21}-\frac{7.4374}{729.125}\) =\(\frac{5}{21}-\frac{42}{125}\) =\(\frac{-257}{2625}\) b)S=\(2^2+4^2+....+20^2\) =\(\left(1.2\right)^2+\left(2.2\right)^2+....+\left(10.2\right)^2\) =\(2^2.\left(2^2+4^2+....+10^2\right)\) =\(2^2.385\) =4.385 =\(1540\)
