Đáp án: a.$ x=\dfrac{12}{17}$
b.$x\in\{-\dfrac12,\dfrac13\}$
c.$x\in\{-\dfrac{55}{168},\dfrac{167}{168}\}$
Giải thích các bước giải:
a.Ta có:
$3\dfrac12-\dfrac4{2x}=\dfrac23$
$\to \dfrac{3\cdot 2+1}2-\dfrac2{x}=\dfrac23$
$\to \dfrac{7}2-\dfrac2{x}=\dfrac23$
$\to \dfrac2{x}=\dfrac{7}2-\dfrac23$
$\to \dfrac2{x}=\dfrac{17}6$
$\to x=\dfrac{2\cdot 6}{17}$
$\to x=\dfrac{12}{17}$
b.Ta có:
$(x+\dfrac12)(\dfrac23-2x)=0$
$\to x+\dfrac12=0\to x=-\dfrac12$
Hoặc $\dfrac23-2x=0\to 2x=\dfrac23\to x=\dfrac13$
c.Ta có:
$\dfrac34-|2x-\dfrac23|=-\dfrac47$
$\to |2x-\dfrac23|=\dfrac34+\dfrac47$
$\to |2x-\dfrac23|=\dfrac{37}{28}$
$\to 2x-\dfrac23=\dfrac{37}{28}$
$\to 2x=\dfrac23+\dfrac{37}{28}$
$\to 2x=\dfrac{167}{84}$
$\to x=\dfrac{167}{168}$
Hoặc $2x-\dfrac23=-\dfrac{37}{28}$
$\to 2x=\dfrac23-\dfrac{37}{28}$
$\to 2x=-\dfrac{55}{84}$
$\to x=-\dfrac{55}{168}$
