$d:x\geqslant -\frac{2}{3} \ \Rightarrow S1=[-\frac{2}{3},\infty )$
$x<-\sqrt{5}\Rightarrow S2=(-\infty,-\sqrt{5} )$
$S1\bigcap S2=\phi $
$d:x\geqslant -\frac{2}{3} \ \Rightarrow S1=[-\frac{2}{3},\infty )$
$x<-\sqrt{5}\Rightarrow S2=(-\infty,-\sqrt{5} )$
$S1\bigcap S2=\phi $
; 12-26-2016 11:00 PM
" "
$16:b)\because \left | x-4 \right |+\left | x+4 \right |=8$
$\therefore x-4\leqslant 0\Lambda x+4\geqslant 0$
$x\leq 4\Lambda x\geqslant -4\Rightarrow -4\leqslant x\leqslant 4$
$S=\left [ -4,4 \right ]$
" "
$16:c)\because \left | x-4 \right |-\left | x+4 \right |=8$
$\therefore x-4\leqslant 0\Lambda x+4\leqslant 0$
$x\leq 4\Lambda x\leq -4\Rightarrow x\leq - 4$
$S=\left (-\infty ,-4 \right ]$
; 12-27-2016 09:50 PM
" "
$16:d)\because \left | x+4 \right |-\left | x-4 \right |=8$
$\therefore x+4\geq 0\Lambda x-4\geq 0$
$x\geq -4\Lambda x\geqslant 4\Rightarrow x\geq 4$
$S=\left [ 4,\infty \right )$
" "
$17)\left | x-3 \right |+\left | x+2 \right |-\left | x-4 \right |=3$
$when:x-3\leqslant 0,x+2\geq 0,x-4\leq 0$
$-x+3+x+2+x-4=3\Rightarrow x=2$
$when:x-3\leqslant 0,x+2\leqslant 0,x-4\leqslant 0$
$-x+3-x-2+x-4=3\Rightarrow -x-3=3\Rightarrow x=-6$
$S=\left \{ 2,-6 \right \}$
" "
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