2021 x 2021 - 2019 x 2023 

= (2019 +2) x ( 2023 -2) - 2019 x 2023

= 2019 x 2023 - 2 x 2019 + 2 x 2023 - 4 - 2019 x 2023

= ( 2019 x 2023 - 2019 x 2023) + 2 x ( 2023 - 2019) - 4

= 0 + 2 x 4 - 4

= 8 - 4

 = 4

2021 x 2021 - 2019 x 2023 

= (2019 +2) x ( 2023 -2) - 2019 x 2023

= 2019 x 2023 - 2 x 2019 + 2 x 2023 - 4 - 2019 x 2023

= ( 2019 x 2023 - 2019 x 2023) + 2 x ( 2023 - 2019) - 4

= 0 + 2 x 4 - 4

= 8 - 4

 = 4

=[2019+2]*[2023-2]-2019*2023

=2019*2023-2*2019+2*2023-4-2019*2023

=[2019*2023-2019*2023]+2*[2023-2019]-4

=0+2*4-4

Mình làm thế này chưa chắc đã đúng đâu nhưng vẵn có thể đúng

A = 2019 \(\times\) 2021 + 2023

A = (2018 + 1).(2022 -1) + 2023 

A = 2018.2022 - 2018 + 2023 > 2018.2022 - 2022

Vậy A > B 

Cách 1: Nhìn qua là biết A > B :)) 

Cách 2: Giải cụ thể:

A = 2019 x 2021 + 2023

   = 2018 x 2021 + 2021 + 2023 = 2018 x 2021 + 4044

B = 2018 x 2022 - 2022

   = 2018 x 2021 + 2018 - 2022 = 2018 x 2021 - 4

⇒ A > B và lớn hơn: 4044 + 4 = 4048

Lời giải:

$A=(-1-2+3+4)+(-5-6+7+8)+(-9-10+11+12)+...+(-2021-2022+2023+2024)-2024$

$=\underbrace{4+4+...+4}_{506}-2024$
$=4.506-2024=0$

2. Tìm x:

( x - 3 )² - x + 3 = 0

=> x² - 6x + 9 - x + 3 = 0

=> x² - 7x + 12 = 0

=> ( x² - 3x ) + ( 4x - 12 ) = 0

=> x.(x - 3) + 4.(x - 3) = 0

=> ( x - 3 ).( x + 4 ) = 0

=> x - 3 = 0 => x = 3

     x + 4 = 0 => x = -4

Trl:

1.

a. \(75^2+150\text{.}25+25^2\)

\(=75^2+2\text{.}75\text{.}25+25^2\)

\(=\left(75+25\right)^2\)

\(=100^2\)

\(=10000\)

b. \(2019^2-2019.19-19^2-19.1981\)

(Đề bài có sai ko vậy???)~ hoặc lak do mk ngu quá k bt lm

2. \(\left(\text{x}-3\right)^2-\text{x}+3=0\)

\(\text{x}^2-6\text{x}+9-\text{x}+3=0\)

\(\text{x}^2-7\text{x}+12=0\)

\(\text{x}^2-3\text{x}-4\text{x}+12=0\)

\(\text{x}\left(\text{x}-3\right)-4\left(\text{x}-3\right)=0\)

\(\left(\text{x}-3\right)\left(\text{x}-4\right)=0\)

\(\orbr{\begin{cases}\text{x}-3=0\\\text{x}-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\text{x}=3\\\text{x}=4\end{cases}}}\)

Vậy ....

#HuyềnAnh#

Bài 1:

a) \(75^2+150.25+25^2\)

\(=75^2+2.75.25+25^2\)

\(=\left(75+25\right)^2\)

\(=100^2\)

\(=10000\)

b) \(2019^2-2019.19-19^2-19.1981\)

\(=\left(2019^2-19^2\right)-2019.19-19.1981\)

\(=\left(2019-19\right).\left(2019+19\right)-19\left(2019+1981\right)\)

\(=2000.2038-19.4000\)

\(=2000.2038-19.2.2000\)

\(=2000.2038-38.2000\)

\(=2000.\left(2038-38\right)\)

\(=2000.2000\)

\(=4000000\)

Bài 2:

\(\left(x-3\right)^2-x+3=0\)

\(\Rightarrow\left(x-3\right)^2-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x-3-1\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}}\)

Vậy \(x\in\left\{3;4\right\}\)

4

4

4

=(1-2)-(3-4)+(5-6)-(7-8)+...+(2021-2022)-2023
=(-1)-(-1)+(-1)-...+(-1)-2023
=0-2023
=-2023

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow\frac{x+1+2019}{2019}+\frac{x+2+2018}{2018}+\frac{x+3+2017}{2017}=\frac{x-1+2021}{2021}+\frac{x-2+2022}{2022}+\frac{x-3+2023}{2023}\)\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)

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

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

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Đặt \(A=\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|+\left|x-2021\right|\)

Ta có: \(\hept{\begin{cases}\left|x-2021\right|=\left|2021-x\right|\\\left|x-2020\right|=\left|2020-x\right|\end{cases}}\)

Ta lại có: \(\hept{\begin{cases}\left|x-2018\right|+\left|2021-x\right|\ge\left|x-2018+2021-x\right|=3\\\left|x-2019\right|+\left|2020-x\right|\ge\left|x-2019+2020-x\right|=1\end{cases}}\)

 \(\Rightarrow\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|+\left|x-2021\right|\ge1+3=4\)

 \(\Rightarrow A_{min}=4\)

Dấu '=' xảy ra khi: \(\hept{\begin{cases}\left(x-2018\right).\left(2021-x\right)\ge0\\\left(x-2019\right).\left(2020-x\right)\ge0\end{cases}}\)

                        \(\Rightarrow\hept{\begin{cases}2018\le x\le2021\\2019\le x\le2020\end{cases}}\)\(\Rightarrow2018\le x\le2020\)

Vậy \(A_{min}=4\)\(\Leftrightarrow\)\(2018\le x\le2020\)

Nếu các bạn chưa hiểu chỗ suy ra ở chỗ dấu bằng xảy ra thì bạn hãy lập bảng xét dấu nhé ^_^

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