(toán 8) tìm x biết

C

chaugiang81

x khác 0
$\dfrac{x-1}{3}= \dfrac{4}{\dfrac{3}{x} +3}$
$\dfrac{x-1}{3}= \dfrac{4}{\dfrac{3x +3}{x}}$
$<=> \dfrac{x-1}{3}= \dfrac{4x}{3(x+1)}$
$<=> x-1= \dfrac{4x}{x+1}$
$<=> x^2 -1= 4x $
$<=> x^2 -4x +4 - 5 =0$
$<=> (x-2)^2= 5$
$=> x-2 = \sqrt{5}$
$=> x= \sqrt{5} +2 $
hoặc $x-2= -\sqrt{5}$
$=>x=2 -\sqrt{5} $
:D:D
 
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