\(=\left[\left(\dfrac{4}{5}\right)^5:\left(\dfrac{2}{5}\right)^6+\dfrac{2^{15}\cdot3^8}{2^9\cdot2^6\cdot3^6}\right]\cdot\dfrac{3^{15}\cdot5^{30}}{3^{20}\cdot5^{30}}\)

\(=\left[\dfrac{4^5}{5^5}\cdot\dfrac{5^6}{2^6}+9\right]\cdot\dfrac{1}{3^5}\)

\(=\dfrac{25}{3^5}=\dfrac{25}{243}\)

MCD:R2//R1

\(R=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{18\cdot12}{18+12}=7,2\left(\Omega\right)\)

\(I=\dfrac{E}{R+r}=\dfrac{12}{7,2+0,8}=1,5\left(A\right)\)

\(U_2=U_1=U=R\cdot I=7,2\cdot1,5=10,8\left(V\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{10,8}{12}=0,9\left(A\right)\)

Đổi 16 phút 5s=965 s

\(m=\dfrac{1}{F}\cdot\dfrac{A}{n}\cdot I_2\cdot t=\dfrac{1}{96500}\cdot\dfrac{108}{1}\cdot0,9\cdot965=0,972\left(kg\right)\)

a) Ta có: \(\widehat{A}=\widehat{B}=65^0\)

Mà 2 góc này đồng vị

=> m//n

b) Ta có: m//n, CD⊥n

=> CD⊥m

c) Ta có: m//n

\(\Rightarrow\widehat{GHD}+\widehat{G}=180^0\)(trong cùng phía)

\(\Rightarrow\widehat{GHD}=180^0-110^0=70^0\)

a)\(\widehat{DBA}=\widehat{CAM}=65^o\) mà 2 góc này đồng vị với nhau ⇒m//n

b) CD⊥n,m//n⇒CD⊥m

c) Ta có m//n \(\Rightarrow\widehat{CGH}+\widehat{DHG}=180^o\) (2 góc trong cùng phía)

                      \(\Rightarrow110^o+\widehat{DHG}=180^o\\ \Rightarrow\widehat{DHG}=70^o\)