\(A=-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-...-\dfrac{1}{5^{99}}+\dfrac{1}{5^{100}}\)

\(=-\dfrac{1}{5}\left(1-\dfrac{1}{5}\right)-\dfrac{1}{5^3}\left(1-\dfrac{1}{5}\right)-...-\dfrac{1}{5^{99}}\left(1-\dfrac{1}{5}\right)\)

\(=\left(1-\dfrac{1}{5}\right)\left(-\dfrac{1}{5}-\dfrac{1}{5^3}-...-\dfrac{1}{5^{99}}\right)\)

\(=\left(\dfrac{1}{5}-1\right)\left(\dfrac{1}{5}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\right)\)

Mặt khác:

\(F=\dfrac{1}{5}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\)

\(25F=5+\dfrac{1}{5}+...+\dfrac{1}{5^{97}}\)

\(25F-F=5-\dfrac{1}{5^{99}}\)

\(F=\dfrac{5-\dfrac{1}{5^{99}}}{24}\)

\(\Rightarrow A=\left(\dfrac{1}{5}-1\right).F\)

\(=\dfrac{-4}{5}.\dfrac{5-\dfrac{1}{5^{99}}}{24}=\dfrac{\dfrac{1}{5^{99}}-5}{5.6}=\dfrac{\dfrac{1}{5^{100}}-1}{6}\)

d) \(A=7-\dfrac{1}{5}+\dfrac{1}{3}-6-\dfrac{9}{5}-\dfrac{4}{3}\)

\(=1-2-1=-2\)

e) \(A=7+\dfrac{7}{12}-\dfrac{1}{2}+3-\dfrac{1}{12}-5\)

\(=5+\dfrac{7-6-1}{12}=5+0=5\)

f) \(\dfrac{13}{23}-\dfrac{15}{4}+\dfrac{10}{23}-\dfrac{1}{4}-\dfrac{2}{27}\)

\(=1-4-\dfrac{2}{27}=-3-\dfrac{2}{27}=-\dfrac{83}{27}\)

g) \(\left(\dfrac{3}{5}\right)^{10}.\left(\dfrac{5}{3}\right)^{10}-\left(\dfrac{13}{39}\right)^4+2014^0\)

\(=1-\dfrac{1}{3^4}+1\)

\(=2-\dfrac{1}{81}=\dfrac{161}{81}\)

\(3^2.5+2^3.10-3^4:3\)

\(=5\left(3^2+2^3.2\right)-3^{4-1}\)

\(=5\left(9+16\right)-3^3\)

\(=5.25-27\)

\(=125-27=98\)

Ta có: 

tam giác AEB = tam giác CFD 

=> \(\widehat{AEB}=\widehat{CFD}=\widehat{EDF}\left(slt\right)\) 

mà 2 goác có vị trí đồng vị

=> EB//DF

Mặt khác: ED//BF

=> EBFD là h.b.h

Ta có: 

Tam giác END= tam giác FMB

=> DN=BM

=> DN+MN=BM+MN=BN

Ta có:

Vì tứ giác ABCD và EBFC đều là h.b.h

=> AC, BD, EF đồng quy tại trung điểm của EF

\(D=\dfrac{4}{5}+\dfrac{4}{5}^2+\dfrac{4}{5}^3+...+\dfrac{4}{5}^{200}\)

\(\dfrac{4}{5}D=\dfrac{4}{5}^2+\dfrac{4}{5}^3+...+\dfrac{4}{5}^{201}\)

\(\dfrac{4}{5}D-D=\dfrac{4}{5}^{201}-\dfrac{4}{5}\)

\(-\dfrac{1}{5}D=\dfrac{4}{5}^{201}-\dfrac{4}{5}\)

\(D=\dfrac{\dfrac{4}{5}-\dfrac{4}{5}^{201}}{\dfrac{1}{5}}=4-\dfrac{4^{201}}{5^{200}}\)

Bạn không hiểu rồi

M nằm trên trục dương mà

0->1 có 6 khoảng

=> 1 khoảng \(=\dfrac{1}{6}\)

Từ 0-> M có 7 khoảng

=> \(M=\dfrac{7}{6}\)

\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)

\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)

\(=\dfrac{1}{2021}\)

a) Tam giác ABE= tam giác CDF

=> EB=DF

b) Ta có: 

\(\widehat{ABE}=\widehat{FCD}\)

\(\Rightarrow\widehat{EDF}=\widehat{EBF}=\widehat{BEA}\)

=> EB//CD mà ED//BF

=> EBFD là h.b.h

c) Gọi K là trung điểm EF

=> K là trung điểm AC, BD, EF

=> AC, BD, EF đồng quy tại K

Câu 1: A

Câu 2: C

Câu 3: B

Câu 4: 0->1 có 6 khoảng

=> 1 khoảng \(=\dfrac{1}{6}\)

=> \(M=\dfrac{7}{6}\)

=> Chọn A