Đáp án:
a.$x\in\{\dfrac43,\dfrac53\}$
b.$ x\in\{\dfrac29,\dfrac53\}$
c.$x\in\{-\dfrac56,\dfrac13\}$
d.$x=\dfrac{53}8$
Giải thích các bước giải:
a.Ta có:
$|2x-3|-\dfrac13=0$
$\to |2x-3|=\dfrac13$
$\to 2x-3=\dfrac13\to 2x=\dfrac{10}3\to x=\dfrac53$
Hoặc $2x-3=-\dfrac13\to 2x=\dfrac83\to x=\dfrac43$
b.Ta có:
$|2x-1|-|x+\dfrac{1}3|=0$
$\to |2x-1|=|x+\dfrac{1}3|$
$\to 2x-1=x+\dfrac{1}3$
$\to 2x-x=1+\dfrac13$
$\to x=\dfrac53$
Hoặc $2x-1=-(x+\dfrac{1}3)$
$\to 2x-1=-x-\dfrac13$
$\to 2x+x=1-\dfrac13$
$\to 3x=\dfrac23$
$\to x=\dfrac29$
c.Ta có:
$\dfrac56-|x+\dfrac14|=\dfrac14$
$\to |x+\dfrac14|=\dfrac56-\dfrac14$
$\to |x+\dfrac14|=\dfrac7{12}$
$\to x+\dfrac14=\dfrac7{12}$
$\to x=\dfrac13$
Hoặc $x+\dfrac14=-\dfrac7{12}\to x=-\dfrac56$
d.Ta có:
$3x-|x+15|=\dfrac54$
$\to 3x=|x+15|+\dfrac54$
Mà $|x+15|+\dfrac54\ge 0+\dfrac54>0$
$\to 3x>0$
$\to x>0$
$\to x+15>0$
$\to |x+15|=x+15$
$\to 3x=(x+15)+\dfrac54$
$\to 3x=x+\dfrac{53}4$
$\to 2x=\dfrac{53}4$
$\to x=\dfrac{53}8$
