\(\Leftrightarrow\dfrac{x-2}{2020}-1+\dfrac{x-3}{2019}-1=\dfrac{x-2019}{3}-1+\dfrac{x-2020}{2}-1\)

=>x-2022=0

hay x=2022

Ta có: \(|2019-x|+|2021-x|=|2019-x|+|x-2021|\)

\(\ge|2019-x+x-2021|=|-2|=2\)

Dấu " = " xảy ra khi \(\left(2019-x\right)\cdot\left(x-2021\right)\ge0\) => 2019 - x và x - 2021 cùng dấu

\(TH1:\hept{\begin{cases}2019-x< 0\\x-2021< 0\end{cases}\Rightarrow\hept{\begin{cases}x>2019\\x< 2021\end{cases}}\Rightarrow2019< x< 2021}\) 

\(TH2:\hept{\begin{cases}2019-x\ge0\\x-2021\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\le2019\\x\ge2021\end{cases}}}\) ( loại )

Mà \(|2019-x|+|2020-x|+|2021-x|=2\)

\(\Rightarrow|2020-x|=0\Rightarrow2020-x=0\Rightarrow x=2020-0=2020\)

Vì 2020 thỏa mãn lớn hơn 2019 và bé hơn 2021 => x = 2020

\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}=-4\)

=> \(\left[\frac{x+1}{2020}+1\right]+\left[\frac{x+2}{2019}+1\right]+\left[\frac{x+3}{2018}+1\right]+\left[\frac{x+4}{2017}+1\right]=-4\)

=> \(\left[\frac{x+1}{2020}+\frac{2020}{2020}\right]+\left[\frac{x+2}{2019}+\frac{2019}{2019}\right]+\left[\frac{x+3}{2018}+\frac{2018}{2018}\right]+\left[\frac{x+4}{2017}+\frac{2017}{2017}\right]=-4\)

=>  \(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}=-4\)

=> \(\left[x+2021\right]\left[\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\right]=-4\)

Do \(\frac{1}{2020}>\frac{1}{2019}>\frac{1}{2018}>\frac{1}{2017}\)nên \(\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\ne0\)

Do đó : x + 2021 = -4 => x = -4 - 2021 = -2025

a: \(\left|x-1\right|\ge0\forall x\)

=>\(\left|x-1\right|+2\ge2\forall x\)

=>\(\left(\left|x-1\right|+2\right)^2\ge4\forall x\)

Ta có: \(\left(\left|x-1\right|+2\right)^2\ge4\forall x\)

\(\left|y-z\right|\ge0\forall y,z\)

Do đó: \(\left(\left|x-1\right|+2\right)^2+\left|y-z\right|\ge4\forall x,y,z\)

=>\(\left(\left|x-1\right|+2\right)^2+\left|y-z\right|+2020\ge4+2020=2024\forall x,y,z\)

Dấu '=' xảy ra khi x-1=0 và y-z=0

=>x=1 và y=z

b: \(A=\left|x-2019\right|-\left|x-2020\right|\)

=>A<=|x-2019-x+2020|=1∀x

Dấu '=' xảy ra khi (x-2019)(x-2020)<=0

=>2019<=x<=2020

<=> (x² +2019x)-(2020x+2019.2020)=0
<=> x.(x+2019)-2020.(x+2019)=0
<=>(x-2020).(x+2019)=0
câu kia tương tự

Sửa đề: \(\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\cdot x=\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

Ta có: \(\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

\(=\left(1+\frac{2019}{2}\right)+\left(1+\frac{2018}{3}\right)+\cdots+\left(1+\frac{1}{2020}\right)+1\)

\(=\frac{2021}{2}+\frac{2021}{3}+\cdots+\frac{2021}{2021}=2021\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\)

Ta có: \(\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\cdot x=\frac{2020}{1}+\frac{2019}{2}+\cdots+\frac{1}{2020}\)

=>\(x\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)=2021\left(\frac12+\frac13+\cdots+\frac{1}{2021}\right)\)

=>x=2021

|x - 2019| + |2019 - x| = 2

⇔ |x - 2019| + |2019 - x| = 2 

⇔ |x - 2019| - |x - 2019| = 2

⇔ 2.|x - 2019| = 2

⇔ |x - 2019| = 1

⇔ x = 2020

⇔ |x - 2019| = -1

⇔ x = 2018

Vạy x = 2020 hoặc x = 2018

Tính [G_(x) - f_(x) ] = ( \(1-x^2+.....+x^{2020}\)) -  (\(x^{2020}-x^{2019}+....-x+1\))

                          = (\(x^{2020}-x^{2019}+....-x+1\)) - (\(x^{2020}-x^{2019}+....-x+1\))

                          = 0

=> h_(x) = [G_(x) - f_(x) ] * [G_(x) + f_(x) ]

            = 0 * [G_(x) + f_(x) ]

           = 0