Ta có:

x+(x+1)+(x+2)+(x+3)+.......+2018+2019=2019x+(x+1)+(x+2)+(x+3)+.......+2018+2019=2019

⇒x+(x+1)+(x+2)+(x+3)+.......+2018=0⇒x+(x+1)+(x+2)+(x+3)+.......+2018=0

Số số hạng là: (Số cuối−Số đầu) : Khoảng cách+1=(2018−x) : 1+1= 2019

Trung bình cộng: (Số đầu+số cuối) : 2=( 2018+x) : 2

Như vậy ta được:

(2019−x).2018+x : 2=0

⇒2019−x=0⇒x=2019 (loại) (vì nếu x=2019 thì số số hạng là 0) hoặc 2018+x=0⇒x=−2018

Vậy x=-2018

Trl

-Bạn kia làm đúng rồi nhé ~!

Chúc bạn học tốt

#Mưaa

\(a,PT\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\Leftrightarrow x^2=-1\left(\text{vô nghiệm}\right)\)

Vậy: ...

\(b,PT\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy: ...

\(c,PT\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

Vậy: ...

\(d,PT\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)

Vậy: ...

\(e,PT\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Vậy: ...

\(f,PT\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\Leftrightarrow x=\pm\dfrac{3}{5}\)

Vậy: ...

khó quá bẹn gì đấy ơi

\(x=0\) không là nghiệm của phương trình

Chia hai vế phương trình cho x, phương trình trở thành:

\(\left(x+\dfrac{4}{x}\right)+2-m=4\sqrt{x+\dfrac{4}{x}}\left(1\right)\)

Đặt \(x+\dfrac{4}{x}=t\left(t\ge2\right)\)

\(\left(1\right)\Leftrightarrow m=f\left(t\right)=t^2-4t+2\left(2\right)\)

Phương trình đã cho có nghiệm khi phương trình \(\left(2\right)\) có nghiệm \(t\ge2\)

\(\Leftrightarrow m\ge f\left(2\right)=-2\)

\(\Rightarrow\) có 2021 giá trị thỏa mãn yêu cầu bài toán

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

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

a) (x-3) + (x-2) + ( x-1) + ..... + 10 + 11 = 11

(x-3) + (x-2) + ( x-1) + ..... + 10 = 0

Gọi số các số hạng từ x-3 đến 10 là n

Ta có : [10 + (x-3)].n : 2 = 0

(x+7).n = 0

Vì n ≠ 0 ( n là số các số hạng )

Nên x+7 = 0

x = 0-7

x = -7

Vậy x = -7

b)

 x + ( x + 1 ) + ( x + 2 ) + ... + 2018 + 2019 = 2019 

⇒ x + ( x +1 ) + ...  + 2018 = 0 

⇒ x + ( x + 1 ) + ... + ( x + 2018 ) = 1 + 2 + ... + 2018 

⇒ x = 0 

vậy x = 0 

cái . ở giữa 2019 . \(\sqrt{x^4}\) là x hay bài khác vậy ?