1/  196

2/  5/4

3/  1/3

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

Ta có: |x - 1| = |1 - x|

Lại có: A = |x - 2019| + |1 - x| ≥ |x - 2019 + 1 - x| = |-2018| = 2018

Dấu " = " xảy ra <=> (x - 2019)(1 - x) ≥ 0

Th1: \(\hept{\begin{cases}x-2019\ge0\\1-x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge2019\\x\le1\end{cases}}\)(Vô lý)

Th2: \(\hept{\begin{cases}x-2019\le0\\1-x\le0\end{cases}}\Rightarrow\hept{\begin{cases}x\le2019\\x\ge1\end{cases}}\Rightarrow1\le x\le2019\)

Vậy GTNN A = 2018 khi 1 ≤ x ≤ 2019

mình ko giúp đc rồi

Ta có: \(A=\left|x-2018\right|+\left|2019-x\right|+\left|x-2020\right|\)

\(A=\left(\left|x-2018\right|+\left|2020-x\right|\right)+\left|2019-x\right|\)

\(\Rightarrow A\ge\left|x-2018+2020-x\right|+\left|2019-x\right|=2+\left|2019-x\right|\)

Dấu "=" xảy ra <=> \(\left(x-2018\right)\left(2020-x\right)\ge0\)

\(\Rightarrow\left(x-2018\right)\left(x-2020\right)\le0\)

\(\Rightarrow\hept{\begin{cases}x-2018\ge0\\x-2020\le0\end{cases}\Rightarrow\hept{\begin{cases}x\ge2018\\x\le2020\end{cases}\Rightarrow}2018\le x\le2020}\)

Và \(\left|2019-x\right|\ge0\), Min (A) = 2 <=> |2019-x| = 0 <=> x= 2019

\(A=\left|2018-x\right|+\left|2019-x\right|+\left|2020-x\right|\)

\(=\left|2018-x\right|+\left|2019-x\right|+\left|x-2020\right|\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) :

\(A\ge\left|2018-x+x-2020\right|+\left|2019-x\right|=2+\left|2019-x\right|\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\left(2018-x\right)\left(x-2020\right)\ge0;2019-x=0\Leftrightarrow x=2019\left(tm\right)\)

Vậy GTNN của A là 2 tại x=2019

\(A=\left(|2018-x|+|2020-x\right)+|2019-x|\)

Đặt \(B=|2018-x|+|2020-x|\)

\(=|2018-x|+|x-2020|\ge|2018-x+x-2020|\)

Hay \(B\ge2\left(1\right)\)

Dấu "=" xảy ra\(\Leftrightarrow\left(2018-x\right)\left(x-2020\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}2018-x\ge0\\x-2020\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}2018-x< 0\\x-2020< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\le2018\\x\ge2020\end{cases}\left(loai\right)}\)hoặc \(\hept{\begin{cases}x>2018\\x< 2020\end{cases}}\)

\(\Leftrightarrow2018< x< 2020\)

Đặt \(C=|2019-x|\)

Vì \(|2019-x|\ge0;\forall x\)

Hay \(C\ge0;\forall x\left(2\right)\)

Dấu "=" xảy ra\(\Leftrightarrow2019-x=0\)

                       \(\Leftrightarrow x=2019\)

Từ (1) và (2) \(\Rightarrow B+C\ge2+0\)

Hay \(A\ge2\)

Dấu "=" xảy ra\(\Leftrightarrow\hept{\begin{cases}2018< x< 2020\\x=2019\end{cases}\Leftrightarrow}x=2019\)

Vậy MIN A=2 \(\Leftrightarrow x=2019\)

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