20222022 x 2023 - 20232023 x 2022

= 10001 x 2022 x 2023 - 10001 x 2023 x 2022

= 0

= 0

-2021 . 20222022 + 2022 .20212021

= -2021 . 2022. 10001 + 2022.2021. 10001

= 0

-2021 . 20222022 + 2022 . 20212021

= (-2021) . (20222022 - 2022 . 10001)

= (-2021) . (20222022 - 20222022)

= (-2021) . 0

= 0

olm sẽ hướng dẫn em làm bài này như sau:

Bước 1: em giải phương trình tìm; \(x\); y

Bước 2:  thay\(x;y\) vào P

(\(x-1\))²⁰²² + |y + 1| = 0

Vì (\(x-1\))²⁰²² ≥ 0 ∀ \(x\); |y + 1| ≥ 0  ∀ y

⇒ (\(x\) - 1)²⁰²²  + |y + 1| = 0

⇔ \(\left\{{}\begin{matrix}\left(x-1\right)^{2022}=0\\y+1=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\) (1) 

Thay (1) vào P ta có:

1²⁰²³.(-1)²⁰²² : )(2.1- 1)²⁰²² +  2023

=  1 + 2023

= 2024

a+b+c=12

   \(\dfrac{2021}{2022}\) x \(\dfrac{2022020222022}{202320232023}\) x \(\dfrac{20212021}{20232023}\)

= \(\dfrac{2021}{2022}\) x \(\dfrac{2022}{2023}\) x \(\dfrac{2021}{2023}\)

= \(\dfrac{2021\times2021}{2023\times2023}\)

= \(\dfrac{4084441}{4092529}\)

Thay \(x=97\) vào

\(=\left(97\right)^2+6.97+2022\\ =9409+582+2022\\ =12013\)

Vậy tại \(x=97\) thì \(P=12013\)

P=x^2+6x+9+2013

=(x+3)^2+2013

=2013+10000

=12013

a) \(M=2022-\left|x-9\right|\le2022\)

\(maxM=2022\Leftrightarrow x=9\)

b) \(N=\left|x-2021\right|+2022\ge2022\)

\(minN=2022\Leftrightarrow x=2021\)

\(M=x^{2023}-2023.\left(x^{2022}-x^{2021}+x^{2020}-x^{2019}+...+x^2-x\right)\)

Ta có : \(x=2022\Rightarrow x+1=2023\)

\(\Rightarrow M=x^{2023}-\left(x+1\right).\left(x^{2022}-x^{2021}+x^{2020}-x^{2019}+...+x^2-x\right)\)

\(\Rightarrow M=x^{2023}-\left(x+1\right)x^{2022}+\left(x+1\right)x^{2021}-\left(x+1\right)x^{2020}+\left(x+1\right)x^{2019}+...-\left(x+1\right)x^2+\left(x+1\right)x\)

\(\Rightarrow M=x^{2023}-x^{2023}-x^{2022}+x^{2022}+x^{2021}-x^{2021}-x^{2020}+x^{2020}+x^{2019}-x^{2019}-...-x^3-x^2+x^2+x\)

\(\Rightarrow M=x\)

\(\Rightarrow M=2022\)

Vậy \(M=2022\left(tạix=2022\right)\)

\(\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(8^2-576:3^2\right)\)

\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-576:3^2\right)\)

\(=\left(1^1+2^2+3^3+4^4+...+2022^{2022}\right)\left(64-64\right)\)

\(=\left(1^1+2^2+3^3+4^4+2022^{2022}\right).0\)

\(=0\)

Ta có :                  

                ^{82 - 576 : 32}

^{= 64 - 576 : 9}

^{= 64 - 64}

^{=  0}

 (1¹ + 2² + 3³ + 4⁴ +...+ 2022²⁰²²) . 0

^{= 0}           

đợi tý

Đã trả lời rồi còn độ tí đồ ngull

\(A=\dfrac{3\cdot\dfrac{a}{b}-\dfrac{-a}{b}}{-\dfrac{-5a}{b}+\dfrac{4a}{b}}\\ =\left(\dfrac{3a}{b}+\dfrac{a}{b}\right):\left(\dfrac{5a}{b}+\dfrac{4a}{b}\right)\\ =\dfrac{4a}{b}:\dfrac{9a}{b}\\ =\dfrac{4a}{b}\cdot\dfrac{b}{9a}\\ =\dfrac{4}{9}\)

Vậy `a=2021/2022` ; `b=2023/2022` thì `A=4/9`

Thanks ạ