Đáp án:
Giải thích các bước giải:
$I=-2|\dfrac{1}{3}x+4|+1\dfrac{2}{3}$
$ $
$=-2|\dfrac{1}{3}x+4|+\dfrac{5}{3}$
$ $
Ta có: $-2|\dfrac{1}{3}x+4|≤0,∀x$
$ $
$⇒I=-2|\dfrac{1}{3}x+4|+1\dfrac{2}{3}≤\dfrac{2}{3}$
$ $
$⇒Imax=\dfrac{2}{3}$ khi $\dfrac{1}{3}x+4=0⇔x=-12$
$ $
$ $
$ $
$ $
$K=\dfrac{3}{2(3x+1)^{6}+3|1-y|^{3}+2}$
$ $
$⇒Kmax$ khi $2(3x+1)^{6}+3|1-y|^{3}+2$ đạt GTNN
$ $
Ta có: $2(3x+1)^{6}≥0,∀x ; 3|1-y|^{3}≥0,∀y$
$⇒2(3x+1)^{6}+3|1-y|^{3}+2≥2$
$⇒2(3x+1)^{6}+3|1-y|^{3}+2$ đạt GTNN là 2 khi $\left \{ {{3x+1=0} \atop {1-y=0}} \right.$ ⇔$\left \{ {{x=\frac{-1}{3}} \atop {y=1}} \right.$
$ $
$⇒Kmax=\dfrac{3}{2}$ khi $\left \{ {{3x+1=0} \atop {1-y=0}} \right.$ ⇔$\left \{ {{x=\frac{-1}{3}} \atop {y=1}} \right.$
