Question

Chứng minh rằng 3^2+3^3+3^4+...+3^98+3^99+3^100 chia hết cho 13

Answer 1

(3^2+3^3+3^4)+...+(3^98+3^99+3^100)=13.3^2+....+13.3^98=13.(3^2+...+3^98)chia het cho 13

 

 

 

 

Answer 2

Đặt $x=\sqrt[3]{3+2\sqrt{2}},y=\sqrt[3]{3-2\sqrt{2}}$
$\Rightarrow \left\{\begin{matrix} x^{3}+y^{3}=6\\xy=1 \end{matrix}\right.$
$\Rightarrow (x+y)^{3}=x^{3}+y^{3}+3xy(x+y)=6+3xy=3[1+1+(x+y)]> 3.3\sqrt[3]{1.1.(x+y)}$
(Vì x>1,y>0=>x+y>1)
Do đó: $(x+y)^{3}> 3^{2}.\sqrt[3]{x+y}$
$\Rightarrow (x+y)^{9}>3^{6}.(x+y)$
$\Rightarrow (x+y)^{8}>3^{6}$
=>đpcm

Answer 3

may cai bai lop 1 re vai

Answer 4

A=1.2+2.3+3.4+.............+2019.2020
3A=1.2.3+2.3.3+3.4.3+........................+2019.2020.3
3A=1.2.3+2.3.(4-1)+3.4.(5-2)+..............+2019.2020.(2021-2018)
3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+.............-2018.2019.2020+2019.2020.2021
3A=2019.2020.2021
A=
3
2019.2020.2021
A=2747468660
Vậy A=2747468660
Chúc bn học tốt

Answer 5

A=1.2+2.3+3.4+.............+2019.2020 3A=1.2.3+2.3.3+3.4.3+........................+2019.2020.3 3A=1.2.3+2.3.(4-1)+3.4.(5-2)+..............+2019.2020.(2021-2018) 3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+.............-2018.2019.2020+2019.2020.2021 3A=2019.2020.2021 A= 3 2019.2020.2021 A=2747468660 Vậy A=2747468660 Chúc bn học tốt