[toán 9] giải phương trình

D

depvazoi

$13\sqrt{x-1}+9\sqrt{x+1}=16x (x \ge 1)$
$<=>16x-13\sqrt{x-1}-9\sqrt{x+1}=0$
$<=>13x-13-13\sqrt{x-1}+\dfrac{13}{4}+3x+3-9\sqrt{x+1}+\dfrac{27}{4}=0$
$<=>13(\sqrt{x-1}-\dfrac{1}{2})^2+3(\sqrt{x+1}-\dfrac{3}{2})^2=0$
$<=> \left\{\begin{matrix} \sqrt{x-1}-\dfrac{1}{2}=0\\\sqrt{x+1}-\dfrac{3}{2}=0 \end{matrix}\right.$
$<=>x=\dfrac{5}{4} (N)$
$Vậy x=\dfrac{5}{4}$
 
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