a.

Ta có: \(m^2+1\ne0;\forall m\Rightarrow\) hàm số là hàm bậc nhất với mọi m

b.

\(m^2+1\ge1>0\) ; \(\forall m\Rightarrow\) hàm đồng biến với mọi m

Ta có: 

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

=> \(\frac{2019}{2020}.a=\frac{2019}{2020}-\left(\frac{2019}{2020}\right)^2+\left(\frac{2019}{2020}\right)^3-...+\left(\frac{2019}{2020}\right)^{2020}-\left(\frac{2019}{2020}\right)^{2021}\)

Lấy

 \(a+\frac{2019}{2020}a=1-\left(\frac{2019}{2020}\right)^{2021}\)

<=> \(a\left(1+\frac{2019}{2020}\right)=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right]\)

<=> \(a.\frac{4039}{2020}=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right]\)

<=> \(a.=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right].\frac{2020}{4039}\)

Vì : \(0< \left(\frac{2019}{2020}\right)^{2021}< 1\)

=> \(0< 1-\left(\frac{2019}{2020}\right)^{2021}< 1\)

và \(0< \frac{2020}{4039}< 1\)

=> \(0< \left[1-\left(\frac{2019}{2020}\right)^{2021}\right].\frac{2020}{4039}< 1\)

=> 0 < a < 1

=> a không phải là một số nguyên.

toan lop may vay ban ?

có A=2(1+2)+2^3(1+2)+..+2^2019(1+2)

       =2.3+2^3.3+...+2^2019.3

       =3(2+2^3+...+2^2019)⋮3

Có A=2(1+2+2^2+2^3+2^4)+...+2^2016(1+2+2^2+2^3+2^4)

          =2.31+.....+2^2016.31

         =31.(2+....+2^2016)⋮31

A=7 mu 2020 mu 2019-3 mu 2016 mu 2015 :5 chung to A la so chan

A=2+2^2+2^3+...+2^2020

= (2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)+.....+ VIẾT LIỀN VÀO (2^2016+2^2017+2^2018+2^2019+2^2020)

=2.(1+2+2^2+2^3+2^4)+2^6.(1+2+2^2+2^3+2^4)+...+ 2^2016.(1+2+2^2+2^3+2^4)

= 2.31+2^6.31+.....+2^2016.31

=31.(2+2^6+...+2^2016) chia hết cho 31

Suy ra A chia hết cho 31

mk mất công làm nên bn tick cho mk nhé

`1/2^2 < 1/(1.2)`

`1/3^3 < 1/(2.3)`

`...`

`1/(2020^2) < 1/(2019.2020)`

`=> A < 1/(1.2) + 1/(2.3) + ... + 1/(2019.2020)`

`=> A < 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/2019 - 1/2020 < 1`.

\(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{2020^2}< \dfrac{1}{2019.2020}\)

Vậy \(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2019.2020}=\dfrac{1}{1}-\dfrac{1}{2020}=\dfrac{2019}{2020}< \dfrac{2020}{2020}=1\)

  S        =  5 + 5² + 5³ +...+ 5²⁰²⁰ + 5²⁰²¹

5S        = 5²+ 5³ + 5⁴ +...+ 5²⁰²¹ + 5²⁰²²

5S-S =(5² + 5³ + 5⁴ + ... + 5²⁰²¹ + 5²⁰²²)-(5 + 5² + 5³ + ... + 5²⁰²¹)

4S   = 5² + 5³ + 5⁴ +...+ 5²⁰²¹ + 5²⁰²² - 5 - 5² - 5³ - ...- 5²⁰²¹

4S   = (5² - 5²)+(5³- 5³)+(5⁴ - 5⁴) + ... +(5²⁰²¹ - 5²⁰²¹)+(5²⁰²² - 5)

4S   = 5²⁰²² - 5

4S + 5 = 5²⁰²² - 5 + 5

4S + 5 = 5²⁰²²  (đpcm)