Giải phương trình vô tỉ

H

huytrandinh

$<=>\sqrt{3x+1}-4+1-\sqrt{6-x}+(3x+1)(x-5)=0$
$<=>\dfrac{3(x-5)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+(3x+1)(x-5)=0$
$.x-5=0<=>x=5$
$.x-5\neq 0$
$<=>\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1=0$
$dk \dfrac{-1}{3}\leq x\leq 6$
$=>VT> 0=>VN$
 
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