x = 4

=>(2x-7)^2017[(2x-7)^2-1]=0

=>(2x-7)(2x-8)(2x-6)=0

hay \(x\in\left\{3;3.5;4\right\}\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-\dfrac{7}{2}\right)\left(2x-\dfrac{5}{2}\right)=0\)

\(\Leftrightarrow x\in\left\{\dfrac{3}{2};\dfrac{7}{4};\dfrac{5}{4}\right\}\)

|2x+2017|=2019

=>\(|^{2x+2017=2019}_{2x+2017=-2019}\)

=>\(|^{2x=2019-2017}_{2x=-2019-2017}\)

=>\(|^{2x=2}_{2x=-4036}\)

=>\(|^{x=2:2}_{x=-4036:2}\)

=>\(|^{x=1}_{x=-2018}\)

Vậy x\(\in\left\{1;-2018\right\}\)

Chúc bn học tốt!

|2x + 2017|= 2019

\(\Rightarrow\)2x + 2017 = 2019 & 2x + 2017 = -2019

TH1: 2x + 2017 = 2019

2x = 2019 - 2017

2x = 2

x = 2 ÷ 2

x = 1

TH2: 2x + 2017 = -2019

2x = -2019 - 2017

2x = - 4036

x = -4036 ÷ 2

x = - 2018

Vây x \(\in\){ 1 ; -2018}

Đi mà hỏi Ông Thắng 

     \(\left(7^{2017}-7^{2018}+7^{2019}\right):7^{2017}\)

\(=7^{2017}\left(1-7+7^2\right):7^{2017}\)

\(=1-7+7^2\)

\(=1-7+49\)

\(=53\)

(7²⁰¹⁷ - 7²⁰¹⁸ + 7²⁰¹⁹) : 7²⁰¹⁷

= 7²⁰¹⁷ : 7²⁰¹⁷ - 7²⁰¹⁸ : 7²⁰¹⁷ + 7²⁰¹⁹ : 7²⁰¹⁷

= 7^{2017 - 2017} - 7^{2018 - 2017} + 7^{2019 - 2017}

= 1 - 7 + 7²

= 1 - 7 + 49

= -6 + 49

= 43

Tham khảo :

Nhận thấy vế trái luôn dương nên \(x-2020\ge0\Leftrightarrow x\ge2020\)

Với \(x\ge2020\Rightarrow\left\{{}\begin{matrix}x-2017\ge0\\2x-2018\ge0\\3x-2019\ge0\end{matrix}\right.\)

PT trở thành: \(x-2017+2x-2018+3x-2019=x-2020\)

Hay kết hợp với điều kiện \(x=\dfrac{4034}{5}\) suy ra PT đã cho vô nghiệm 

Đặt \(2017-x=m,2019-x=n\)

\(\rightarrow m+n=2x-4036\)

Phương trình ban đầu trở thành :

\(m^3+n^3=\left(m+n\right)^3\)

\(\rightarrow3mn.\left(m+n\right)^3=0\)

\(\rightarrow\left(2017-x\right)\left(2019-x\right)\left(2x-4036\right)=0\)

\(\rightarrow\left[{}\begin{matrix}x=2017\\x=2018\\x=2019\end{matrix}\right.\)

Vậy \(S=\left\{2017;2018;2019\right\}\)

(2017-X)³+(2019-X)³+(2X-4036)³=0

<=>(2017-x).(2018-x).(2019-x)=0

<=>x=2017

x=2018

x=2019

#YQ

\(\Leftrightarrow\left(2x-1\right)\left(...\right)=0\Rightarrow x=\frac{1}{2}\)

\(\frac{2x-1}{2020}-\frac{2x-1}{2019}+\frac{2x-1}{2018}=\frac{2x-1}{2017}-\frac{2x-1}{2016}\\ \Leftrightarrow\frac{2x-1}{2020}-\frac{2x-1}{2019}+\frac{2x-1}{2018}-\frac{2x-1}{2017}+\frac{2x-1}{2016}=0\\ \Leftrightarrow\left(2x-1\right)\left(\frac{1}{2020}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}+\frac{1}{2016}\right)=0\)

mà \(\frac{1}{2020}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}+\frac{1}{2016}\ne0\)

thì \(2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\frac{1}{2}\)

vậy \(x=\frac{1}{2}\)