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`#3107`

\(\left(3^{2021}+3^{2020}\right)\div3^{2020}\\ =3^{2021}\div3^{2020}+3^{2020}\div3^{2020}\\ =3^{2021-2020}+3^{2020-2020}\\ =3+1=4\)

giúp đê

\(...\dfrac{3^{2021}+3^{2020}}{3^{2020}}\\ =\dfrac{3^{2019}.\left(3^2+3\right)}{3^{2019}.3}\\ =\dfrac{3^2+3}{3}\\ =\dfrac{9+3}{3}\\ =\dfrac{12}{3}\\ =4.\)

ghi lại đề:

12.x - 33 = 3² . 3³

=> 12.x - 33 = 243

=> 12x = 243 + 33

     12x = 276

=> x = 276 : 12

     x = 23

chúc bạn học tốt!! ^^

ok mk nhé!!! 356535645746857579763563653645645654645657657657858753453465465756765

12x - 33 = 3².3³

12x - 33 = 3⁵

12x - 33 = 243

12x = 243+33

12x = 276

x= 276:12

x = 23

vậy x = 23

\(12x-33=3^2.3^3\)

\(12x-33=3^5\)

\(12x-33=243\)

\(12x=243+33\)

\(12x=276\)

\(x=276\div12\)

\(x=23\)

cảm ơn các bạn đã chỉ mình nha, thanks nhiều

\(\Leftrightarrow3A=3^2+3^3+...+3^{2022}\\ \Leftrightarrow3A-A=3^2+3^3+...+3^{2022}-3-3^2-...-3^{2021}\\ \Leftrightarrow2A=3^{2022}-3\\ \Leftrightarrow A=\dfrac{3^{2022}-3}{2}\)

\(=\dfrac{3^{2022}-3}{2}\)

12x-33=3².3³

12x-33=3²⁺³

12x-33=3⁵

12x-33=243

12x=243+33

12x=276

x=276:12

x=23

tk nhé

12x - 33 = 3^2 x 3^3

12x - 33 = 3^2+3

12x - 33 = 3^5

12x - 33 = 243

12x       = 243 + 33

12x       =   276

    x       = 276 : 12

    x       =     23

\(12x-33=3^2\times3^3\)

\(12x-33=243\)

\(12x=243+33\)

\(12x=276\)

\(x=276:12\)

\(x=23\)
 

A=[1+3+3^2+3^3]+...+[3^2018+3^2019+3^2020+3^2021]

A=1 nhân[1+3+3^2+3^3]+...+3^2018 nhân [1+3+3^2+3^3]

A=[1+3+3^2+3^3] NHÂN[1+...+3^2018

A=40 nhân [1+...+3^2018]

=> A chia hết cho 40

\(A=1+3^1+3^2+3^3+...+3^{2021}\\=(1+3^1)+(3^2+3^3)+(3^4+3^5)...+(3^{2020}+3^{2021})\\=4+3^2\cdot(1+3)+3^4\cdot(1+3)+...+3^{2020}\cdot(1+3)\\=4+3^2\cdot4+3^4\cdot4+...+3^{2020}\cdot4\\=4\cdot(1+3^2+3^4+...+3^{2020})\)

Vì \(4\cdot(1+3^2+3^4+...+3^{2020})\vdots4\)

nên \(A\vdots4\)

\(\text{#}Toru\)

thank you bạn character debate nha, ai vô trả lời thì cảm ơn nhiều!!