Đáp án:b.$B\le\dfrac13$
c.$C\le 6$
Giải thích các bước giải:
b.Ta có:
$(x-2.3)^2\ge 0,\quad\forall x$
$\to (x-2.3)^2+3\ge 0+3,\quad\forall x$
$\to\dfrac{1}{(x-2.3)^2+3}\le\dfrac13$
$\to B\le\dfrac13$
Dấu = xảy ra khi $(x-2.3)^2=0\to x=2.3$
c.Ta có:
$C=\dfrac{34+15|x+2|}{6|x+2|+8}$
$\to C=\dfrac{14+(15|x+2|+20)}{2(3|x+2|+4)}$
$\to C=\dfrac{14+5(3|x+2|+4)}{2(3|x+2|+4)}$
$\to C=\dfrac{14}{2(3|x+2|+4)}+\dfrac52$
Mà $|x+2|\ge 0$
$\to 3|x+2|+4\ge 3\cdot 0+4=4$
$\to \dfrac{14}{2(3|x+2|+4)}\le\dfrac{14}{4}=\dfrac72$
$\to \dfrac{14}{2(3|x+2|+4)}+\dfrac52\le \dfrac72+\dfrac52$
$\to C\le 6$
Dấu = xảy ra khi $|x+2|=0\to x=-2$
