Giải thích các bước giải:
Bài 1:
a.Ta có:
$\dfrac15-4.2+\dfrac{12}7-\dfrac23-\dfrac57-\dfrac13$
$=0.2-4.2+\dfrac{12}7-\dfrac57-\dfrac23-\dfrac13$
$=-(-0.2+4.2)+(\dfrac{12}7-\dfrac57)-(\dfrac23+\dfrac13)$
$=-4+\dfrac{12-5}7-\dfrac{2+1}3$
$=-4+\dfrac77-\dfrac33$
$=-4+1-1$
$=-4$
b.Ta có:
$(\dfrac32)^3\cdot 4-\dfrac{17}2\cdot\sqrt{36}+\dfrac1{\sqrt{4}}$
$=\dfrac{3^3}{2^3}\cdot 4-\dfrac{17}2\cdot 6+\dfrac1{2}$
$=\dfrac{27}{8}\cdot 4-51+\dfrac1{2}$
$=\dfrac{27}{2}-51+\dfrac1{2}$
$=-51+\dfrac{27}{2}+\dfrac1{2}$
$=-51+\dfrac{27+1}{2}$
$=-51+\dfrac{28}{2}$
$=-51+14$
$=-(51-14)$
$=-37$
Câu 2:
a.Ta có:
$\dfrac34+\dfrac14:x=\dfrac52$
$\to\dfrac14:x=\dfrac52-\dfrac34$
$\to \dfrac14:x=\dfrac74$
$\to x=\dfrac14:\dfrac74$
$\to x=\dfrac17$
b.Ta có:
$|2x-1|+\dfrac13=\dfrac43$
$\to |2x-1|=\dfrac43-\dfrac13$
$\to |2x-1|=1$
$\to 2x-1=1\to 2x=2\to x=1$
Hoặc $2x-1=-1\to 2x=0\to x=0$
