\(A=1+\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{2020}\)

=> \(3A=3+1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{2019}\)

=> \(3A-A=\left(3+1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{2019}\right)-\left(1+\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{2020}\right)\)

=> \(2A=3-\frac{1}{3^{2020}}=\frac{3^{2021}-1}{3^{2020}}\)

=> \(A=\frac{3^{2021}-1}{3^{2020}.2}\)

a,b,c,d,e = 0

g = 3/10

máy tính xin hân hạnh đồng hành cùng với lê Kiều Trinh

x=8,25

Cách giải sao ạ ??

Ns rõ đề bài hơn đi

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\left(a+b-c\right)^2=a^2+b^2+c^2+2ab-2bc-2ac\)

\(\left(a-b-c\right)^2=a^2+b^2+c^2-2av-2ac+2bc\)

Chúc bn hoc tốt

a) \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)

\(=7^2-4.12=49-48=1\)

b(\(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

\(=20^2+4.3=400+12=412\)

Cm: a, Ta có:

(a+b)² = a² + 2ab +b² (1)

(a-b)² + 4ab = a² - 2ab +b² + 4ab = a² + 2ab +b² ( 2)

Từ (1), (2) => đpcm

b. Ta có 

(a-b)² = a² - 2ab +b²  (3)

(a+b)² - 4ab = a² + 2ab +b² - 4ab = a² - 2ab +b² (4)

Từ (3),(4)=> đpcm

Áp dụng tính chất:

a, (a-b)² = (a+b)² - 4ab = 7² -4.12 = 1

b,(a+b)² = (a-b)² + 4ab = 20² + 4.3 = 412

Chúc bn hc tốt!

\(A=\frac{8\frac{3}{9}.5\frac{1}{4}+3\frac{16}{19}.5\frac{1}{4}}{\left(2\frac{14}{17}-2\frac{1}{34}\right).34}:\frac{7}{24}\)

\(=\frac{5\frac{1}{4}\left(8\frac{1}{3}+3\frac{16}{19}\right)}{\left(\frac{28}{34}-\frac{1}{34}\right).34}:\frac{7}{24}\)

\(=\frac{\frac{21}{4}\left(\frac{25}{3}+\frac{73}{19}\right)}{\frac{27}{34}.34}:\frac{7}{24}\)

\(=\frac{\frac{21}{4}\left(\frac{475}{57}+\frac{219}{57}\right)}{27}:\frac{7}{24}\)

\(=\frac{\frac{21}{4}.\frac{674}{57}}{27}:\frac{7}{24}\)

\(=\frac{\frac{14154}{228}}{27}.\frac{24}{7}=\frac{\frac{339696}{228}}{189}\)

Ban sai r Famas a

475+219=694 chu