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`
