Bạn thông cảm , máy mình mất kí hiệu "không chia hết " rồi

Ta có : (x,6) = 1 mà 6 ⋮ 2 => x không chia hết cho 2

=> x lẻ => x² là số chính phương lẻ

+) Ta có : x² là số chính phương lẻ => x² chia 8 dư 1

=> x² - 1 ⋮ 8 (1)

+) Cmtt => x không chia hết chó 3

=> x² chia 3 dư 1 => x²-1 ⋮ 3 (2)

Từ (1) , (2) , (3,8)=1 , 3.8 = 24

=> x ⋮ 24

Sửa đề : \(\dfrac{x+7}{2015}+\dfrac{x+8}{2016}+\dfrac{x+9}{2017}+\dfrac{x+10}{2018}+x+\dfrac{11}{2019}=5\)

Ta có :

\(\dfrac{x+7}{2015}+\dfrac{x+8}{2016}+\dfrac{x+9}{2017}+\dfrac{x+10}{2018}+x+\dfrac{11}{2019}=5\)

\(\Rightarrow\left(\dfrac{x+7}{2015}-1\right)+\left(\dfrac{x+8}{2016}-1\right)+\left(\dfrac{x+9}{2017}-1\right)+\left(\dfrac{x+10}{2018}-1\right)+\left(\dfrac{x+11}{2019}-1\right)=0\)

\(\Rightarrow\dfrac{x-2008}{2015}+\dfrac{x-2008}{2016}+\dfrac{x-2008}{2017}+\dfrac{x-2008}{2018}+\dfrac{x-2008}{2019}=0\)

\(\Rightarrow\left(x-2008\right).\left(\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}+\dfrac{1}{2019}\right)=0\)

mà \(\Rightarrow\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}+\dfrac{1}{2019}\ne0\)

=> x-2008 = 0 => x = 2008

................................

A =    (x-1)(x-3)(x² - 4x + 50)

= (x² - 3x -x +3)(x² - 4x + 50 )

=(x² - 4x + 3) ( x² - 4x + 3 + 47)

= (x² - 4x + 3)(x² - 4x + 3) + 47(x² -4x+3)

= (x² -4x + 3) + 2.\(\dfrac{47}{2}\) (x² -4x + 3) + ( \(\dfrac{47}{2}\))²  - ( \(\dfrac{47}{2}\))²

= (x² - 4x +3 + 47/2 ) ²  -  2209/4

= (x² - 4x + 53/2)² - 2209/4

 vì  x² - 4x + 4 + 45/2 = ( x - 2)² + 45/2 ≥ 45/2

⇔ (x² - 4x + 53/2)² -  2209/4 ≥ (45/2)² - 2209/4 = -46 

⇔A(min) =  - 46 ⇔ x = 2

a) Để \(x=\dfrac{a-4}{a}∈Z\) thì a-4 ⋮ a ( a ≠ 0 )

Vì a - 4 ⋮ a mà a ⋮ a => 4 ⋮ a

=> a ∈ Ư(4)

b) Để x < 0 thì \(x=\dfrac{a-4}{a}< 0\) => a-4 và a khác dấu

Mà a > a - 4 ∀ a

\(\Rightarrow\left\{{}\begin{matrix}a>0\\a-4< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a>0\\a< 4\end{matrix}\right.\) mà a ∈ Ư(4) => a ∈ { 1 ; 2 ; 4 }

Sửa đề : \(\dfrac{2x}{3y}=\dfrac{-1}{3}\) và 2x+3y=7

Ta có : \(\dfrac{2x}{3y}=\dfrac{-1}{3}\Rightarrow\dfrac{2x}{-1}=\dfrac{3y}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\dfrac{2x}{-1}=\dfrac{3y}{3}=\dfrac{2x+3y}{-1+3}=\dfrac{7}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}2x=-1.\dfrac{7}{2}\\3y=3.\dfrac{7}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=\dfrac{-7}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{4}\\y=\dfrac{7}{2}\end{matrix}\right.\)

Nhầm :

Dấu "=" xảy ra \(\left\{{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

(x+1)²+(y-2)¹⁰=0

Vì \(\left\{{}\begin{matrix}\left(x+1\right)^2≥0∀x\\\left(y-2\right)^{10}≥0∀y\end{matrix}\right.\)\(\Rightarrow\left(x+1\right)^2+\ge\left(y-2\right)^{10}0∀x,y\)

Dáu "=" xảy ra <=> \(\left[{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

`(x+7).(x+15)=0`

`=>x+7=0`

`x=0-7`

`x=-7`

`=>x+15=0`

`x=0-15`

`x=-15`

(x+7).(x+15)=0

 \(\Rightarrow\left[{}\begin{matrix}x+7=0\\x+15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=-15\end{matrix}\right.\)

Ta có :

(x-20).(x-8).(x-2022) = 0

\(\Rightarrow\left[{}\begin{matrix}x-20=0\\x-8=0\\x-2022=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=20\\x=8\\x=2022\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=20\\x=8\\x=2022\end{matrix}\right.\)

`(x-20).(x-8).(x-2022)=0`

`=>` \(\left[{}\begin{matrix}x-20=0\\x-8=0\\x-2022=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0+20=20\\x=0+8=8\\x=0+2022=2022\end{matrix}\right.\)

Vậy `x = {20;8;2022}`