$a.A=\bigg(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+.........+\dfrac{1}{44.49}\bigg).\dfrac{1-3-5-7-......-49}{89} \\Đặt\ M=\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+.........+\dfrac{1}{44.49} \\⇒5M=5.\bigg(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+.........+\dfrac{1}{44.49}\bigg) \\⇔5M=\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+.........+\dfrac{5}{44.49} \\⇔5M=\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+.........+\dfrac{1}{44}-\dfrac{1}{49} \\⇔5M=\dfrac{1}{4}-\dfrac{1}{49} \\⇔5M=\dfrac{45}{196} \\⇒M=\dfrac{9}{196} \\Đặt\ N=\dfrac{1-3-5-7-......-49}{89} \\⇔N=\dfrac{1-(3+5+......+49)}{89} \\Số\ số\ hạng là :  (49-3):2+1=24(số) \\⇒N=\dfrac{1-(49+3).24:2}{89}=\dfrac{-623}{89} \\⇒A=M.N=\dfrac{9}{196}.\dfrac{-623}{89}=\dfrac{-9}{28} \\ b.B=\dfrac{2^{12}.3^5-4^6.9^2}{(2^2.3)^6+8^4,3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{(125.7)^3+5^9.14^3} \\=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^6} \\=\dfrac{2^12.3^4(3-1)}{2^{12}.3^5.(3+1)}-\dfrac{5^{10}.7^3(1-7)}{5^9.7^3(1+2^3)} \\=\dfrac{2}{3.4}-\dfrac{5(-6)}{9} \\=\dfrac{1}{6}-\dfrac{-10}{3} \\=\dfrac{7}{2}$

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