\(3+3^2+.....+3^{99}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\)

\(=39+3^3\left(3+3^2+3^3\right)+........+3^{96}\left(3+3^2+3^3\right)\)

\(=39+3^3\cdot39+...+3^{96}\cdot39\)

\(=39\left(1+3^3+....+3^{96}\right)\)

Vì \(39⋮13\Rightarrow39\in B\left(13\right)\)

B(13) là sao bạn

2,65+(- 15,02)+2,3=-10.07

HT

TL

2,65+(-15,02)+2,3=(-12,37)+2,3

                            =-10,07

nha bn

HT

chiu

ối giồi ôi đang học zoom mà còn đăng bài dc

de co phai la 2.(x-25)-17=25

neu de nhu vay thi de lam:

    2x(x-25)-17=25

    2x(x-25)=25+17

    2x(x-25)=42

        x-25 =42:2

        x-25 =42

            x  =42+25

            x  =67

Từ đề bài, ta suy ra:

\(\frac{2020\left(1+2021\right)}{2019\left(2019+3\right)}=\frac{2020\cdot2022}{2019\cdot2022}=\frac{2020}{2019}\)

\(\frac{2020+2020.2021}{2019^2+2019.3}=\frac{2020.\left(1+2021\right)}{2019.\left(2019+3\right)}=\frac{2020.2022}{2019.2022}=\frac{2020}{2019}=1\frac{1}{2019}\)

Chúc bn học tốt

Ta có \(\frac{900}{899}=1+\frac{1}{899}\)

\(\frac{899}{898}=1+\frac{1}{898}\)

Nhận thấy \(\frac{1}{898}>\frac{1}{899}\)

=> \(1+\frac{1}{898}>\frac{1}{899}+1\)

=> \(\frac{899}{898}>\frac{900}{899}\)

Ta có: \(\frac{900}{899}\)=\(\frac{899+1}{899}\)= 1 + \(\frac{1}{899}\)

           \(\frac{899}{898}\)=\(\frac{898+1}{898}\)= 1 + \(\frac{1}{898}\)

Vì 899 > 898

nên \(\frac{1}{899}\)<\(\frac{1}{898}\)

\(\Rightarrow\)1 + \(\frac{1}{899}\)< 1 +\(\frac{1}{898}\)

\(\Rightarrow\)\(\frac{900}{899}\)<\(\frac{899}{898}\)

Vậy ​​​\(\frac{900}{899}\)<\(\frac{899}{898}\).​

không biết

\(\frac{2}{3}=\frac{x-1}{6}\Leftrightarrow3x-3=12\Leftrightarrow3x=15\Rightarrow x=5\)

\(\frac{2}{3}=\frac{x-1}{6}\)

\(\Rightarrow2.6=3.\left(x-1\right)\)

       \(12=3x-3\)

  \(3+12=3x\)

           \(15=3x\)

       \(15:3=x\)

              \(5=x\)

Vậy x=5

Bài nào em ??

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