a/ \(|5x-20|\ge 0→A\ge -1\\→\min A=-1\)

\(→\) Dấu "=" xảy ra khi \(5x-20=0\)

\(↔5x=20\)

\(↔x=4\)

Vậy \(\min A=-1\) khi \(x=4\)

b/ \(|x+5|\ge 0→-|x+5|\le 0→B\le -3\)

\(→\max B=-3\)

\(→\) Dấu "=" xảy ra khi \(x+5=0\)

\(↔x=-5\)

Vậy \(\max B=-3\) khi \(x=-5\)

c/ \(\begin{cases}(x+2)^2\ge 0\\|4y-16|\ge 0\end{cases}\\→(x+2)^2+|4y-16|\ge 0\\→C\ge 3\\→\min C=3\)

\(→\) Dấu "=" xảy ra khi \(\begin{cases}x+2=0\\4y-16=0\end{cases}\)

\(↔\begin{cases}x=-2\\4y=16\end{cases}\)

\(↔\begin{cases}x=-2\\y=4\end{cases}\)

Vậy \(\min C=-3\) khi \(x=-2\) và \(y=4\)

d/ \(\begin{cases}(y-1)^4\ge 0\\|x+3|\ge 0\end{cases}↔\begin{cases}-(y-1)^4\le 0\\-|x+3|\le 0\end{cases}\\→-(y-1)^4-|x+3|\le 0\\→D\le -5\\→\max D=-5\)

\(→\) Dấu "=" xảy ra khi \(\begin{cases}y-1=0\\x+3=0\end{cases}\)

\(↔\begin{cases}y=1\\x=-3\end{cases}\)

Vậy \(\max D=-5\) khi \(x=-3\) và \(y=1\)

e/ Áp dụng BĐT \( |a|+|b|\ge |a+b|\)

\(→|x-2|+|x-6|=|x-2|+|6-x|\ge |x-2+6-x|=|4|=4\)

\(→\min E=4\)

\(→\) Dấu "=" xảy ra khi \( (x-2)(6-x)\ge 0\)

\(↔\left[\begin{array}{1}\begin{cases}x-2\ge 0\\6-x\ge 0\end{cases}\\\begin{cases}x-2\le 0\\6-x\le 0\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}x\ge 2\\x\le 6\end{cases}\\\begin{cases}x\le 2\\x\ge 6\end{cases}(loại)\end{array}\right.\\↔2\le x\le 6\)

Vậy \(\min E=4\) khi \(2\le x\le 6\)

f/ Áp dụng BĐT \(|a|+|b|\ge |a+b|\)

\(→|x-1|+|x-7|=|x-1|+|7-x|\ge |x-1+7-x|=|6|=6\)

\(→\min |x-1|+|x-7|=6\)

\(→\) Dấu "=" xảy ra khi \((x-1)(7-x)\ge 0\)

\(↔\left[\begin{array}{1}\begin{cases}x-1\ge 0\\7-x\ge 0\end{cases}\\\begin{cases}x-1\le 0\\7-x\le 0\end{cases}\end{array}\right.\\↔\left[\begin{array}{1}\begin{cases}x\ge 1\\x\le 7\end{cases}\\\begin{cases}x\le 1\\x\ge 7\end{cases}(loại)\end{array}\right.\\↔1\le x\le 7\)

\(|8-2x|\ge 0→\min |8-2x|=0\)

\(→\) Dấu "=" xảy ra khi \(8-2x=0\)

\(↔2x=8\)

\(↔x=4\)

Ta có: \(F=|x-1|+|x-7|+|8-2x|\\→\min F=6+0=6\)

\(→\) Dấu "=" xảy ra khi \(\begin{cases}1\le x\le 7\\x=4\end{cases}\)

\(→x=4\)

Vậy \(\min F=6\) khi \(x=4\)

Đáp án:

`A = |5x - 20| - 1`

Vì `|5x - 20|` $\geqslant$ `0 ∀ x`

`-> |5x - 20| - 1` $\geqslant$ `1`

`-> A` $\geqslant$ `1`

`-> A_{min} = 1`

Dấu "`=`" xảy ra khi :

`⇔ 5x  -20=0⇔5x=20⇔x=4`

Vậy `A_{min} = 1 ⇔ x = 4`

$\\$

`B = - |x + 5| - 3`

Vì `|x + 5|` $\geqslant$ `0 ∀ x`

`-> - |x + 5|` $\leqslant$ `0 ∀ x`

`-> - |x + 5| - 3` $\leqslant$ `-3`

`-> B` $\leqslant$ `-3`

`-> B_{max} = -3`

Dấu "`=`" xảy ra khi :

`⇔ x + 5 = 0 ⇔ x = -5` 

Vậy `B_{max} = -3⇔ x = -5`

$\\$

`C = (x + 2)^2 + |4y - 16| + 3`

Vì \(\left\{ \begin{array}{l}(x + 2)^2 \geqslant 0∀x\\|4y-16|\geqslant 0∀y\end{array} \right.\)

`-> (x + 2)^2 + |4y - 16|` $\geqslant$ `0 ∀ x,y`

`-> (x + 2)^2 + |4y - 16| + 3` $\geqslant$ `3`

`-> C` $\geqslant$ `3`

`-> C_{min} = 3`

Dấu "`=`" xảy ra khi :

`⇔` \(\left\{ \begin{array}{l}x+2=0\\4y-16=0\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}x=-2\\y=4\end{array} \right.\)

Vậy `C_{min} = 3 ⇔ x=-2,y=4`

$\\$

`D = - (y - 1)^4 - |x + 3| -5`

Vì \(\left\{ \begin{array}{l}(y-1)^4\geqslant 0∀y\\|x+3|\geqslant 0 ∀x\end{array} \right.\)

`->` \(\left\{ \begin{array}{l}-(y-1)^4\leqslant0∀y\\-|x+3|\leqslant0∀x\end{array} \right.\)

`-> - (y - 1)^4 - |x + 3|` $\leqslant$ `0 ∀ x,y`

`-> - (y - 1)^4 - |x + 3| - 5` $\leqslant$ `-5`

`-> D` $\leqslant$ `-5`

`-> D_{max} = -5`

Dấu "`=`" xảy ra khi :

`⇔` \(\left\{ \begin{array}{l}y-1=0\\x+3=0\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}y=1\\x=-3\end{array} \right.\)

Vậy `D_{max} = -5 ⇔ y = 1, x = -3`

$\\$

`E = |x - 2| + |x - 6|`

`-> E = |x - 2| + |6 - x|`

Áp dụng BĐT `|a| + |b|` $\geqslant$  `|a + b|`

`-> |x - 2| + |6 - x|` $\geqslant$ `|x - 2 + 6 - x| = |4| = 4`

`-> E` $\geqslant$ `4`

`->E_{min} = 4`

Dấu "`=`" xảy ra khi :

`(x - 2) (6 - x)` $\leqslant$ `0`

`⇔` \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x-2\geqslant 0\\6-x\geqslant0\end{array} \right.\\ \left\{ \begin{array}{l}x-2\leqslant0\\6-x\leqslant 0\end{array} \right.\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}\left\{ \begin{array}{l}x\geqslant 2\\x\leqslant6\end{array} \right.\text{(Luôn đúng)}\\ \left\{ \begin{array}{l}x\leqslant2\\x \geqslant6\end{array} \right. \text{(Vô lí)}\end{array} \right.\)

`⇔ 2` $\leqslant$ `x` $\leqslant$ `6`

Vậy `E_{min} = 4 ⇔ 2` $\leqslant$ `x` $\leqslant$ `6`

$\\$

`F = |x - 1| + |x - 7| + |8 - 2x|`

`-> F = |x - 1| + |7 - x| +|8 - 2x|`

Áp dụng BĐT `|a| + |b|` $\geqslant$ `|a + b|`

`-> |x - 1| + |7 - x| + |8 - 2x|` $\geqslant$ `|x - 1+ 7 - x| + |8 - 2x|`

`-> |x - 1| + |7 - x| + |8 - 2x|` $\geqslant$ `6 + |8 - 2x|`

Vì `|8 - 2x|` $\geqslant$`∀ x`

`-> 6 + |8 - 2x|`$ \geqslant$ `6`

`-> F` $\geqslant$ `6`

`-> F_{min} = 6`

Dấu "`=`" xảy ra khi :

\(\left\{ \begin{array}{l}(x - 1) (7 - x)\leqslant0 \\8 - 2x =0\end{array} \right.\) `⇔` \(\left\{ \begin{array}{l}(x - 1) (7 - x)\leqslant0 \\x=4\end{array} \right.\)

Bạn tự giải `(x - 1) (7 - x)` $\leqslant$ `0`

`-> 1` $\leqslant$ `x` $\leqslant$ `7`

`-> x = 4`

Vậy `F_{min} = 6 ⇔ x = 4`