Toán Lớp 8: 40đ:V
Tìm x
$\dfrac{2x+\dfrac{2(3-x)}{5}}{12}$=$1+\dfrac{1-\dfrac{9-2x}{12}}{5}$
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Toán Lớp 8: 40đ:V
Tìm x
$\dfrac{2x+\dfrac{2(3-x)}{5}}{12}$=$1+\dfrac{1-\dfrac{9-2x}{12}}{5}$
Comments ( 2 )
⇔$\dfrac{2x+\dfrac{2\left(3-x\right)}{5}}{12}=1+\dfrac{1-\dfrac{9-2x}{12}}{5}$
$⇔\dfrac{2x+\dfrac{2\left(3-x\right)}{5}}{12}\cdot \:60=1\cdot \:60+\dfrac{1-\dfrac{9-2x}{12}}{5}\cdot \:60$
$⇔2\left(4x+3\right)=60+12-\left(9-2x\right)$
$⇔8x+6=2x+63$
$⇔8x+6-6=2x+63-6$
$⇔8x=2x+57$
$⇔8x-2x=2x+57-2x$
$⇔6x=57$
$⇔\dfrac{6x}{6}=\dfrac{57}{6}$
$⇔x=\dfrac{19}{2}$
Vậy $x=\dfrac{19}{2}$
$\dfrac{2x+\dfrac{2\left(3-x\right)}{5}}{12}=1+\dfrac{1-\dfrac{9-2x}{12}}{5}$
$⇒\dfrac{2x+\dfrac{2\left(3-x\right)}{5}}{12}\cdot \:60=1\cdot \:60+\dfrac{1-\dfrac{9-2x}{12}}{5}\cdot \:60$
$⇒2\left(4x+3\right)=60+12-\left(9-2x\right)$
$⇒8x+6=2x+63$
$⇒8x+6-6=2x+63-6$
$⇒8x=2x+57$
$⇒8x-2x=2x+57-2x$
$⇒6x=57$
$⇒\dfrac{6x}{6}=\dfrac{57}{6}$
$⇒x=\dfrac{19}{2}$
Vậy $x=\dfrac{19}{2}$