a) Ta có:

Suy ra: 

Do đó .

Lại có .

Vậy A < B.

Đây dang lớp mấy vây

A<B

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

Bài 2:

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

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

A < \(\frac{2022^{2022}+\left(1+2021\right)}{2022^{2023}+\left(1+2021\right)}\)

A < \(\frac{2022^{2022}+2022}{2022^{2023}+2022}\)

A < \(\) \(\frac{2022.\left(2022^{2021}+1\right)}{2022.\left(2022^{2022}+1\right)}\)

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

Vậy A < B

Ta có: \(\frac{2022}{1}+\frac{2021}{2}+\cdots+\frac{1}{2022}\)

\(=\left(\frac{2021}{2}+1\right)+\left(\frac{2020}{3}+1\right)+\cdots+\left(\frac{1}{2022}+1\right)+1\)

\(=\frac{2023}{2}+\frac{2023}{3}+\cdots+\frac{2023}{2023}=2023\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)\)

Ta có: \(\frac{\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)}{\frac{2022}{1}+\frac{2021}{2}+\cdots+\frac{1}{2022}}\)

\(=\frac{\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)}{2023\left(\frac12+\frac13+\cdots+\frac{1}{2023}\right)}\)

\(=\frac{1}{2023}\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)

(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))

vậy x= 2023

Ta có; \(B=1-\frac12+\frac13-\frac14+\cdots-\frac{1}{2022}+\frac{1}{2023}\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2023}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\ldots+\frac{1}{2023}-1-\frac12-\cdots-\frac{1}{1011}=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2023}\)

=C

=>B-C=0

A = \(\dfrac{\dfrac{2022}{1}+\dfrac{2021}{2}+\dfrac{2020}{3}+...+\dfrac{1}{2022}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}}\)

Xét TS = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) \(\dfrac{2020}{3}\) +... + \(\dfrac{1}{2022}\)

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

      TS = \(\dfrac{2023}{2}\) + \(\dfrac{2023}{3}\) +...+ \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2023}\)

      TS =  2023.(\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) +...+ \(\dfrac{1}{2023}\))

A = \(\dfrac{2023.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}\)

 A = 2023

Em cảm ơn ạ