Ta có:

\(\widehat{AOC}+\widehat{BOD}=98^o+74^o\)

\(\widehat{AOD}+2\widehat{COD}+\widehat{BOC}=172^o\)

\(\widehat{COD}+\widehat{AOB}=172^o\)

\(\widehat{COD}=172^o-\widehat{AOB}=172^o-136^o=36^o\)

\(3^{n+1}-2.3^n+2^{n+5}-7.2^n\)

\(=3^n\left(3-2\right)+2^n\left(2^5-7\right)\)

\(=3^n+2^n.25\)

Vì \(3^n⋮̸25\); \(25.2^n⋮25\)

=> \(3^n+2^n.25⋮̸25\)

=> \(3^{n+1}-2.3^n+2^{n+5}-7.2^n⋮̸25\)

\(313^5.299-313^6.36\)

\(=313^5.299-313^636\)

\(=313^5\left(299-313.36\right)\)

Ta có:

Ta có: \(299\equiv5\left(mod7\right)\)

       \(313\equiv5\left(mod7\right)\)

       \(36\equiv1\left(mod7\right)\)

=> \(299-313.36\equiv5-5.1=0\left(mod7\right)\)

=> \(299-313.36⋮7\)

=> \(313^5.299-313^6.36⋮7\)

\(A=-\dfrac{1}{3}+\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\)

\(=\dfrac{1}{3}\left(-1+\dfrac{1}{3}\right)+\dfrac{1}{3^3}\left(-1+\dfrac{1}{3}\right)+...+\dfrac{1}{3^{99}}\left(-1+\dfrac{1}{3}\right)\)

\(=\dfrac{-2}{3}\left(\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)\)

Ta có:

\(B=\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

\(9B=3+\dfrac{1}{3}+...+\dfrac{1}{3^{97}}\)

\(9B-B=3-\dfrac{1}{3^{99}}\)

\(B=\dfrac{3-\dfrac{1}{3^{99}}}{8}\)

\(A=-\dfrac{2}{3}B=\dfrac{-2}{3}.\dfrac{3-\dfrac{1}{99}}{8}=\dfrac{\dfrac{1}{3^{100}}-1}{4}\)

\(\left(-0,25\right)^4\cdot4^4\)

\(=\left(-\dfrac{1}{4}\right)^4\cdot4^4\)

\(=\left(-\dfrac{1}{4}\cdot4\right)^4\)

\(=\left(-\dfrac{4}{4}\right)^4\)

\(=\left(-1\right)^4\)

\(=1\)

(-0,25)⁴.4⁴

= (-0,25 . 4)⁴

= (-1)⁴

= 1

Ta có: 

BM là phân giác góc ABC

=> \(\widehat{ABM}=\dfrac{1}{2}\widehat{ABC}\)

Mặt khác:

\(\widehat{ABC}=\widehat{DBE}\) (đối đỉnh)

=> \(\widehat{ABM}=\dfrac{1}{2}\widehat{DBE}\)

\(D=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)

\(=5\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)

\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)

\(=5\left(1-\dfrac{1}{31}\right)\)

\(=5.\dfrac{30}{31}=\dfrac{150}{31}\)