Toán Lớp 8: Tìm x |x|+|x+2|=3 b ) |x-5|=|x+2|

a/ $\begin{array}{|c|cc|}\hline x&&-\infty&&-2&&0&&+\infty\\\hline x&&&-&|&-&0&+&\\\hline x+2&&&-&0&+&|&+\\\hline\end{array}$

Xét khoảng $x<-2$

$→-x-x-2=3\\↔-2x=5\\↔x=-\dfrac{5}{2}(TM)$

Xét khoảng $-2\le x\le 0 $

$\to -x+x+2=3\\↔2=3(vô\,\,lý)$

$→$ Pt vô nghiệm trong khoảng $-2\le x\le 0$

Xét khoảng $x>0$

$→x+x+2=3\\↔2x=1\\↔x=\dfrac{1}{2}(TM)$

Vậy $S=\{-\dfrac{5}{2};\dfrac{1}{2}\}$

c/ $|x-5|=|x+2|\\↔\left[\begin{array}{1}x-5=x+2\\x-5=2-x\end{array}\right.\\↔\left[\begin{array}{1}0=7(vô\,\,lý)\\2x=7\end{array}\right.\\↔x=\dfrac{7}{2}$

Vậy $S=\{\dfrac{7}{2}\}$