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$A=1+2-3-4+5+6-7-8+....+2017+2018-2019-2020+2021+2022$

$=(1+2-3-4)+(5+6-7-8)+...+(2017+2018-2019-2020)+4043$
$=(-4)+(-4)+(-4)+...+(-4)+4043$
Số lần xuất hiện của -4 là: $[(2020-1):1+1]:4=505$

$A=(-4)\times 505+4043=2023$

Câu b có vẻ đề sai. Bạn xem lại nhé.

a: =(-1)+(-1)+...+(-1)=-1011

b: =(-5)+(-5)+...+(-5)=-175

a: \(98^{10}\cdot A=\dfrac{98^{98}+98^{10}}{98^{98}+1}=1+\dfrac{98^{10}-1}{98^{98}+1}\)

\(98^{10}\cdot B=\dfrac{98^{99}+98^{10}}{98^{99}+1}=1+\dfrac{98^{10}-1}{98^{99}+1}\)

98^88+1>98^99+1

=>A<B

b: \(\dfrac{1}{2022^2}\cdot C=\dfrac{2022^{2023}+1}{2022^{2023}+2022^2}=1+\dfrac{1-2022^2}{2022^{2023}+2022^2}\)

\(\dfrac{1}{2022^2}\cdot D=\dfrac{2022^{2021}+1}{2022^{2021}+2022^2}=1+\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

2022^2023>2022^2021

=>2022^2023+2022^2>2022^2021+2022^2

=>\(\dfrac{2022^2-1}{2022^{2023}+2022^2}< \dfrac{2022^2-1}{2022^{2021}+2022^2}\)

=>\(\dfrac{1-2022^2}{2022^{2023}+2022^2}>\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

=>C>D

a) \(S=1+2+3+...+2021\)

\(=\left(2021+1\right).2021:2\)

\(=2043231\)

b) \(P=1+3+5+...+2021\)

\(=\left(2021+1\right).[\left(2021-1\right):2+1]:2\)

\(=2022.1011:2\)

\(=1022121\)

a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.

a: M=-2021+2021-68-68+17

=-119

b: B=(-1)+(-1)+...+(-1)

=-1x500

=-500

c: C=(1-2-3+4)+(5-6-7+8)+...+(997-998-999+1000)

=0

a: \(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35-12}{20}=\dfrac{23}{20}\)

d: \(\left(-\dfrac{1}{4}\right)^2\cdot\dfrac{4}{11}+\dfrac{7}{11}\cdot\left(-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

\(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35}{20}+\dfrac{-12}{20}=\dfrac{23}{20}\)

Lời giải:

$A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2021}}$

$2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2020}}$

$\Rightarrow 2A-A=1-\frac{1}{2^{2021}}$

$\Rightarrow A=1-\frac{1}{2^{2021}}

$B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{60}=\frac{4}{5}=1-\frac{1}{5}$

Hiển nhiên $\frac{1}{2^{2021}}< \frac{1}{5}\Rightarrow 1-\frac{1}{2^{2021}}> 1-\frac{1}{5}$

$\Rightarrow A> B$