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

tìm ? chữ số tận cùng vaz bạn

9 ban nhe

\(\dfrac{x+1}{2023}+\dfrac{x+2}{2022}=\dfrac{x+3}{2021}+\dfrac{x+4}{2020}\\ \Leftrightarrow\dfrac{x+1}{2023}+1+\dfrac{x+2}{2022}+1=\dfrac{x+3}{2021}+1+\dfrac{x+4}{2020}+1\\ \Leftrightarrow\dfrac{x+1+2023}{2023}+\dfrac{x+2+2022}{2022}-\dfrac{x+3+2021}{2021}-\dfrac{x+4+2020}{2020}=0\\ \Leftrightarrow\left(x+2024\right)\times\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)=0\\ \Rightarrow x+2024=0:\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)\\ \Rightarrow x+2024=0\\ \Rightarrow x=-2024\)

Tham khảo câu trả lời:

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

96 nha

nho k

\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)

\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)

Vì \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)

=> x + 2020 = 0

=> x = -2020

            Bài làm :

Ta có :

\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)

\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)

 \(\text{Vì : }\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)

\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)

Vậy x=-2020

\(B=\dfrac{\dfrac{2ab}{3}-\dfrac{3ab}{2}}{-\dfrac{5bb}{6}}\)

\(=\dfrac{\dfrac{4ab}{6}-\dfrac{9ab}{6}}{-\dfrac{5bb}{6}}\)

\(=\dfrac{-\dfrac{5ab}{6}}{-\dfrac{5bb}{6}}=\dfrac{ab.\dfrac{5}{6}}{bb.\dfrac{5}{6}}\)

\(=\dfrac{ab}{bb}=\dfrac{a}{b}\)

Với \(a=\dfrac{2021}{2022};b=\dfrac{2023}{2022}\), ta được:

\(B=\dfrac{2021}{2022}:\dfrac{2023}{2022}=\dfrac{2021}{2022}.\dfrac{2022}{2023}=\dfrac{2021}{2023}\)

Thanks ạ

1

Ta có :

\(B=3^{2021}.7^{2022}.13^{2023}=\left(3^{2020}.3\right).\left(7^{2020}.7^2\right).\left(13^{2023}.13^3\right)\)

\(=\left(3^4\right)^{505}.3.\left(7^4\right)^{505}.7^2.\left(13^4\right)^{505}.13^3\)

= (.......1).3 . (......1).49 . (.......1).(....7)

= (.........3).(......9).(.......7) = (......9)

Vậy chữ số hàng đơn vị của B là 9.