Ta có:

`(3x-6)^2022>=0` với mọi x

`(5y+10)^2024>=0` với mọi y

`=>(3x-6)^2022+(5y+10)^2024>=0` với mọi x,y

Mặt khác: `(3x-6)^2022+(5y+10)^2024)<=0` với mọi x,y

Dấu "=" xảy ra: `3x-6=0` và `5y+10=0`

`<=>3x=6` và `5y=-10`

`<=>x=6/3=2` và `y=-10/5=-2` 

\(A=1^2-2^2+3^2-4^2+...+2023^2-2024^2+2025^2\\ =\left(1+2\right)\left(1-2\right)+\left(3+4\right)\left(3-4\right)+\left(2023-2024\right)\left(2023+2024\right)\\ =-3-7-11-...-4047+2025^2\\ =-\left(3+7+11+..+4047\right)+2025^2\)

Xét tổng: `3+7+11+...+4047`

Số lượng số hạng: `(4047-3):4+1=1012` 

Tổng: `(4047+3)*1012/2=2049300` 

`=>A=-2049300+2025^2`

`=>A=-2049300+4100625`

`=>A=2051325` 

\(\left(3x-3\right)^2+\left(4y+2\right)^2=0\)

Ta có:

`(3x-3)^2>=0` với mọi x

`(4y+2)^2>=0` với mọi x

`=>(3x-3)^2+(4y+2)^2>=0` với mọi x,y

Mặt khác: `(3x-3)^2+(4y+2)^2=0`

Dấu "=" xảy ra: `3x-3=0` và `4y+2=0`

`=>3x=3` và `4y=-2`

`=>x=3/3=1` và `y=-2/4=-1/2` 

Bài 5: 

\(a,\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\\ =\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\\ =\dfrac{24}{24}-\dfrac{41}{41}+\dfrac{1}{2}\\ =1-1+\dfrac{1}{2}\\=\dfrac{1}{2}\\ b,16\cdot\dfrac{3}{5}\cdot\dfrac{-1}{3}-13\dfrac{3}{5}\cdot\dfrac{-1}{3}\\ =\dfrac{-1}{3}\cdot\left(16+\dfrac{3}{5}-13-\dfrac{3}{5}\right)\\ =\dfrac{-1}{3}\cdot3\\ =-1\)

Bài 2:

\(\left(\dfrac{-4}{7}\right)^{25}:\left(\dfrac{-4}{7}\right)^{23}\\ =\left(\dfrac{-4}{7}\right)^{25-23}\\ =\left(\dfrac{-4}{7}\right)^2\\ =\dfrac{16}{49}\\ b,\dfrac{15}{60}+\dfrac{12}{19}+\dfrac{2}{9}-\dfrac{10}{8}+\dfrac{-31}{19}\\ =\dfrac{1}{4}+\left(\dfrac{12}{19}-\dfrac{31}{19}\right)+\dfrac{2}{9}-\dfrac{5}{4}\\ =\left(\dfrac{1}{4}-\dfrac{5}{4}\right)+\dfrac{-19}{19}+\dfrac{2}{9}\\ =-1-1+\dfrac{2}{9}\\ =\dfrac{2}{9}-2\\ =-\dfrac{16}{9}\)

Do n ϵ Z ⇒ A ϵ Z.

\(A=\dfrac{2\left(n-1\right)+5}{n-1}\)

\(A=2+\dfrac{5}{n-1}\)

\(\Rightarrow5⋮\left(n-1\right)\)

\(\Rightarrow\left(n-1\right)\in\left\{1;-1;5;-5\right\}\)

Ta có bảng giá trị:

⇒ Để A đạt GTNN thì A = -3 → n = 0

                   GTLN thì A = 7 → n = 2

`1,` Ta có:

`(4x^3-3x^2y^2+y^5)-(x^3+4x^2y^2+2y^5-5)`

`= 4x^3-3x^2y^2+y^5-x^3-4x^2y^2-2y^5 + 5`

`= (4x^3-x^3)+(-3x^2^2-4x^2y^2)+(y^5-2y^5)+5`

`= 3x^3 - 7x^2y^2 - y^5 + 5`

`2,` Ta có:

`x^2 - 2y^2 + 3z^2 - B = 3x^2 + 2y^2 - z^2`

$\Rightarrow $`(x^2-2y^2+3z^2)-B = 3x^2+2y^2-z^2`

$\Rightarrow $`B = x^2-2y^2+3z^2 - (3x^2 + 2y^2-z^2)`

$\Rightarrow $`B= x^2-2y^2+3z^2-3x^2-2y^2+z^2`

$\Rightarrow $` B = (x^2-3x^2)+(-2y^2-2y^2)+(3z^2+z^2)`

$\Rightarrow $`B = -2x^2-4y^2+4z^2`

$\Rightarrow $ `B`

`(2x + 1)^2 = 1/4`

`=> (2x+1)^2 = (1/2)^2`

`=> 2x + 1 = 1/2` hoặc `2x + 1 = -1/2`

`=> 2x = 1/2 - 1` hoặc `2x = -1/2 - 1`

`=> 2x = -1/2` hoặc `2x = -3/2`

`=> x = -1/2 : 2` hoặc `x = -3/2 : 2`

`=>  x = -1/2 .1/2` hoặc `x = -3/2 . 1/2`

`=> x = -1/4` hoặc `x = -3/4`

Vậy ...

TH1: \(\left(2x+1\right)^2\)\(=\left(\dfrac{1}{2}\right)^2\)       TH2: \(\left(2x+1\right)^2=\left(\dfrac{-1}{2}\right)^2\)

Suy ra: \(2x+1=\dfrac{1}{2}\)              Suy ra: \(2x+1=\dfrac{-1}{2}\)

\(2x=\dfrac{1}{2}-1\)                           \(2x=\dfrac{-1}{2}-1\)

\(2x=\dfrac{-1}{2}\)                               \(2x=\dfrac{-3}{2}\)

\(x=\dfrac{-1}{2}:2\)                            \(x=\dfrac{-3}{2}:2\)

\(x=\dfrac{-1}{2}\cdot\dfrac{1}{2}\)                           \(x=\dfrac{-3}{2}\cdot\dfrac{1}{2}\)

\(x=\dfrac{-1}{4}\)                                 \(x=\dfrac{-3}{4}\)

Vậy \(x\in\left\{\dfrac{-3}{4}|\dfrac{-1}{4}\right\}\)

\(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}-\dfrac{x+21}{2023}-\dfrac{x+20}{2024}=0\)

=>\(\left(\dfrac{x+23}{2021}+1\right)+\left(\dfrac{x+22}{2022}+1\right)-\left(\dfrac{x+21}{2023}+1\right)-\left(\dfrac{x+20}{2024}+1\right)=0\)

=>\(\dfrac{x+2044}{2021}+\dfrac{x+2044}{2022}-\dfrac{x+2044}{2023}-\dfrac{x+2044}{2024}=0\)

=>x+2044=0

=>x=-2044

`(x+23)/2021 + (x+22)/2022 - (x+21)/(2023) - (x+20)/2024 = 0`

`=> (x+23)/2021 + 1+ (x+22)/2022 +1 - (x+21)/(2023) - 1 - (x+20)/2024  - 1= 0`

`=> ((x+23)/2021 + 1)+ ((x+22)/2022 +1) - ((x+21)/(2023) + 1) - ((x+20)/2024  + 1)= 0`

`=> (x+23+2021)/2021 + (x+22+2022)/2022 - (x+21+2023)/(2023) - (x+20+2024)/2024 = 0`

`=> (x+2044)/2021 + (x+2044)/2022 -(x+2044)/(2023) - (x+2044)/2024 = 0`

`=> (x+2044) . (1/2021 + 1/2022 - 1/2023 - 1/2024) = 0`

`=> x + 2044 = 0`

`=> x = -2044`

Vậy `x = -2044`