C với D mình làm sau vì nó phức tạp hơn ... E với F trước nhé

E = | 3x + 1 | + 2| x - y | + 1

\(\hept{\begin{cases}\left|3x+1\right|\ge0\\2\left|x-y\right|\ge0\end{cases}\forall}x,y\Rightarrow\left|3x+1\right|+2\left|x-y\right|+1\ge1\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}3x+1=0\\x-y=0\end{cases}}\Leftrightarrow x=y=-\frac{1}{3}\)

=> MinE = 1 <=> x = y = -1/3

F = 5| x - 1 | + 1/2| 2x + y | + 2020

\(\hept{\begin{cases}5\left|x-1\right|\ge0\\\frac{1}{2}\left|2x+y\right|\ge0\end{cases}\forall}x,y\Rightarrow5\left|x-1\right|+\frac{1}{2}\left|2x+y\right|+2020\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\2x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)

=> MinF = 2020 <=> x = 1 ; y = -2

C = 2| x - 1 | + | 2x + 3 | - 2020

= | 2x - 2 | + | 2x + 3 | - 2020

= | 2x - 2 | + | -( 2x + 3 ) | - 2020

= | 2x - 2 | + | -2x - 3 | - 2020

Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :

C = | 2x - 2 | + | -2x - 3 | - 2020 ≥ | 2x - 2 - 2x - 3 | - 2020 = | -5 | - 2020 = 5 - 2020 = -2015

Dấu "=" xảy ra khi ab ≥ 0

=> ( 2x - 2 )( -2x - 3 ) ≥ 0

Xét hai trường hợp :

1. \(\hept{\begin{cases}2x-2\ge0\\-2x-3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ge2\\-2x\ge3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le-\frac{3}{2}\end{cases}}\)( loại )

2. \(\hept{\begin{cases}2x-2\le0\\-2x-3\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\le2\\-2x\le3\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge-\frac{3}{2}\end{cases}}\Leftrightarrow-\frac{3}{2}\le x\le1\)

=> MinC = -2015 <=> \(-\frac{3}{2}\le x\le1\)

D = | 3 - 2x | + 2| 1 - x | + 1/2

= | 3 - 2x | + | 2 - 2x | + 1/2

= | -( 3 - 2x ) | + | 2 - 2x | + 1/2

= | 2x - 3 | + | 2 - 2x | + 1/2

Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :

D = | 2x - 3 | + | 2 - 2x | + 1/2 ≥ | 2x - 3 + 2 - 2x | + 1/2 = | -1 | + 1/2 = 1 + 1/2 = 3/2

Dấu "=" xảy ra khi ab ≥ 0

=> ( 2x - 3 )( 2 - 2x ) ≥ 0

Xét hai trường hợp :

1. \(\hept{\begin{cases}2x-3\ge0\\2-2x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\ge3\\-2x\ge-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le1\end{cases}}\)( loại )

2. \(\hept{\begin{cases}2x-3\le0\\2-2x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x\le3\\-2x\le-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x\ge1\end{cases}}\Leftrightarrow1\le x\le\frac{3}{2}\)

=> MinD = 3/2 <=> \(1\le x\le\frac{3}{2}\)

Cảm ơn bạn nhiều

Bạn xem lại. Biểu thức C có GTLN chứ không có GTNN bạn nhé.

Áp dụng BĐT Bunyakovsky ta được:

\(\left(x+y\right)\left(\frac{2020}{x}+\frac{1}{2020y}\right)\ge\left(\sqrt{x}\cdot\sqrt{\frac{2020}{x}}+\sqrt{y}\cdot\sqrt{\frac{1}{2020y}}\right)\)

\(=\left(\sqrt{2020}+\sqrt{\frac{1}{2020}}\right)^2=2020+\frac{1}{2020}+2=2022\frac{1}{2020}\)

\(\Leftrightarrow\frac{2021}{2020}\cdot S\ge2022\frac{1}{2020}\)

\(\Rightarrow S\ge2022\frac{1}{2020}\div\frac{2021}{2020}=2021\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{\sqrt{x}}{\sqrt{\frac{2020}{x}}}=\frac{\sqrt{y}}{\sqrt{\frac{1}{2020y}}}\\x+y=\frac{2021}{2020}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2020y\\x+y=\frac{2021}{2020}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2020}\end{cases}}\)

Vậy Min(S) = 2021 khi \(\hept{\begin{cases}x=1\\y=\frac{1}{2020}\end{cases}}\)

\(B=\frac{\left(x-2\right)^2+2016}{\left(x-1\right)^2}=\frac{\left(t-1\right)^2+2016}{t^2}=\frac{t^2-2t+2017}{t^2}\)

\(=1-\frac{2}{t}+\frac{2017}{t^2}=1-2a+2017a^2=2017\left(a^2-2.\frac{1}{4034}+\frac{1}{4034}^2\right)-\frac{2017}{4034^2}+1\)

\(=2017\left(a-\frac{1}{4034}\right)^2+1-\frac{1}{2017^3}\ge1-\frac{1}{2017^3}\)

tự xét dấu = 

\(B=\frac{\left(x-2\right)^2+2016}{\left(x-1\right)^2}\)

\(\Leftrightarrow\frac{\left(t-1\right)^2+2016}{1^2}\)

\(\Leftrightarrow\frac{t^2-2t+2017}{t^2}\)

\(\Leftrightarrow1-\frac{2}{t}+\frac{2017}{t^2}\)

\(\Leftrightarrow1-2a+2017a^2\)

\(\Leftrightarrow a^2-2\times[\frac{1}{4034}+\frac{1^2}{4034}]-\frac{2017}{4034^2}+1\)

\(\Leftrightarrow2017\left(a-\frac{1}{4034}\right)^2+1-\frac{1}{2017}^3\)

phần cuối tự làm nha

TL:

C=\(\frac{2020}{-\left(x^2+2x-2020\right)}\) 

 =\(\frac{2020}{-\left(x^2+2x+1-2021\right)}=\frac{2020}{-\left(x+1\right)^2+2021}\) 

Để Cmin thì \(-\left(x+1\right)^2+2021\) lớn nhất

vì \(-\left(x+1\right)^2+2021\le2021\) =>-(x+1)+2021 lớn nhất =2021

vậy C_min=\(\frac{2020}{2021}\)

\(A=x^2+2x+1+2019\)

\(A=\left(x+1\right)^2+2019\ge2019\)

dấu = xảy ra khi x+1=0

=> x=-1

Vậy...