2021^1=2021 đk ạ??

sigma

S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)

= \(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)

S=P nhé

Ta có: \(S=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2021}+\frac{1}{2022}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\cdots+\frac{1}{2022}-1-\frac12-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}=P\)

=>S-P=0

=>\(\left(S-P\right)^{2022}=0\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022

B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\) 

B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\) 

B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))

Vậy B > C 

Đây nhé bé

Câu1

Vì \(\mid x \mid \geq 0 \Rightarrow \mid x \mid + 1 \geq 1\).
Do đó \(\left(\right. \mid x \mid + 1 \left.\right)^{10} \geq 1^{10} = 1\).

Suy ra:

\(A = \left(\right. \mid x \mid + 1 \left.\right)^{10} + 2023 \geq 1 + 2023 = 2024.\)

Dấu “=” chỉ xảy ra khi \(\mid x \mid = 0 \Leftrightarrow x = 0\).

\(\Rightarrow\) Giá trị nhỏ nhất của \(A\) là \(\boxed{2024}\), đạt tại \(x = 0\).

Câu 2 ( câu này kiến thức nâng cao nhé em nên là khi em đọc lời giải sẽ có khó hiểu nhé )

Đặt \(n = 2022\). Khi đó:

\(A = \frac{n^{2022} + 1}{n^{2023} + 1} , B = \frac{n^{2021} + 1}{n^{2022} + 1} .\)

Xét tổng quát với \(a_{k} = \frac{n^{k} + 1}{n^{k + 1} + 1} , \left(\right. n > 1 \left.\right)\).

Ta gọi k là luỹ thừa của cơ số

\(a_{k} > a_{k - 1} \textrm{ }\textrm{ } \Longleftrightarrow \textrm{ }\textrm{ } \left(\right. n^{k} + 1 \left.\right)^{2} > \left(\right. n^{k + 1} + 1 \left.\right) \left(\right. n^{k - 1} + 1 \left.\right) .\)

Xét hiệu:

\(\left(\right.n^{k}+1\left.\right)^2-\left(\right.n^{k+1}+1\left.\right)\left(\right.n^{k-1}+1\left.\right)=-n^{k-1}\left(\right.n-1\left.\right)^2<0\)

Vậy \(a_{k} < a_{k - 1}\), tức dãy \(\left(\right. a_{k} \left.\right)\) giảm dần theo \(k\)

Do đó:

\(A = a_{2022} < a_{2021} = B .\)

\(\Rightarrow B>A\)

Câu3

Ta đổi : \(27 = 3^{3}\), \(9 = 3^{2}\), \(125 = 5^{3}\).

\(\frac{5^{16} \cdot \left(\right. 3^{3} \left.\right)^{7}}{\left(\right. 5^{3} \left.\right)^{5} \cdot \left(\right. 3^{2} \left.\right)^{11}} = \frac{5^{16} \cdot 3^{21}}{5^{15} \cdot 3^{22}} = 5^{16 - 15} \cdot 3^{21 - 22} = \frac{5}{3} .\)

Vậy kết quả bằng \(\frac{5}{3}\).

Câu 3:

\(\frac{5^{16}\cdot27^7}{125^5\cdot9^{11}}\)

\(=\frac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}=\frac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}\)

\(=\frac53\)

Câu 2:

\(2022A=\frac{2022^{2023}+2022}{2022^{2023}+1}=1+\frac{2021}{2022^{2023}+1}\)

\(2022B=\frac{2022^{2022}+2022}{2022^{2022}+1}=1+\frac{2021}{2022^{2022}+1}\)

Ta có: \(2022^{2023}+1>2022^{2022}+1\)

=>\(\frac{2021}{2022^{2023}+1}<\frac{2021}{2022^{2022}+1}\)

=>\(\frac{2021}{2022^{2023}+1}+1<\frac{2021}{2022^{2022}+1}+1\)

=>2022A<2022B

=>A<B

Câu 1:

\(\left|x\right|\ge0\forall x\)

=>\(\left|x\right|+1\ge1\forall x\)

=>\(\left(\left|x\right|+1\right)^{10}\ge1^{10}=1\forall x\)

=>\(\left(\left|x\right|+1\right)^{10}+2023\ge1+2023=2024\forall x\)

Dấu '=' xảy ra khi x=0

\(...=2022+2020+\left(-2019+2016-2018+2015-2017+2014\right)+...+\left(6-3+5-2+4-1\right)\)

\(=2022+2020+\left(-3-3-3\right)+\left(-3-3-3\right)+...+\left(-3-3-3\right)+\left(-3-2-1\right)\)

\(=2022+2020+\left(-9\right)+\left(-9\right)+...\left(-9\right)+\left(-6\right)\)

\(=2022+2020+\left(-9\right).\left[\left(2019-9\right):6+1\right].\left[\left(2019+6\right)\right]:2+\left(-6\right)\)

\(=2022+2020+\left(-9\right).336.2025:2+\left(-6\right)\)

\(=2022+2020-3061800-6\)

\(=-3057764\)

A=(2011+2019)+(2012+2018)+(2013+2017)+(2014+2016)+2015

A=    4030        +      4030       +         4030    +         4030   +2015

A=                    8060               +                       8060            +2015

 A=                                       17120                                      +2015

A=19135

Giả sử tất cả các số đã cho đều lẻ

=>Quy đồng, ta được:

\(A=\dfrac{\left(a_2\cdot a_3\cdot...\cdot a_{2022}\right)+\left(a_1\cdot a_3\cdot...\cdot a_{2021}\cdot a_{2022}\right)+...+\left(a_1\cdot a_2\cdot...\cdot a_{2021}\right)}{a_1\cdot a_2\cdot...\cdot a_{2022}}=1\)

Tử có 2022 số hạng, mẫu là số lẻ

=>A là số chẵn khác 1

=>Trái GT

=>Phải có ít nhất 1 số là số chẵn

S = 1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²

⇒ S/3 = 1/3² + 1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³

⇒ 2S/3 = S - S/3

= (1/3 + 1/3² + 1/3³ + ... + 1/3²⁰²¹ + 1/3²⁰²²) - (1/3² +1/3³ + 1/3⁴ + ... + 1/3²⁰²² + 1/3²⁰²³)

= 1/3 - 1/3²⁰²³

⇒ S = (1/3 - 1/3²⁰²³) : 2/3

= (1 - 1/3²⁰²²) : 2

Lại có: 1 - 1/3²⁰²² < 1

⇒ S < 1/2