[Toán 10] $\sqrt{2x+1}+\sqrt{x-3}=4\sqrt{9-2x} $

H

huytrandinh

bài này khỏi bình phương chi cho mệt ta có
$pt<=>(\sqrt{2x+1}-3\sqrt{9-2x})+(\sqrt{x-3}-\sqrt{9-2x})=0$
$<=>\dfrac{20(x-4)}{\sqrt{2x+1}+3\sqrt{9-2x}}+\dfrac{3(x-4)}{\sqrt{x-3}+\sqrt{9-2x}}=0$
$.x-4=0<=>x=4$
$.x-4\neq 0$
$=>\dfrac{20}{\sqrt{2x+1}+3\sqrt{9-2x}}+\dfrac{3}{\sqrt{x-3}+\sqrt{9-2x}}=0$
$VT> 0=>VN$
 
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